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ECLECTIC    EDUCATIONAL    SERIES. 


NEW  ELEMENTARY  ALGEBRA, 


PRIMARY   ELEMENTS 


OF 


ALGEBRA, 


COMMON    SCHOOLS    AND    ACADEMIES. 


By   JOSEPH    RAY,    M.  D., 

LATE   PROFESSOR    OF    MATHEMATICS    IN    WOODWARD    COLLEGE. 


REVISED    ELECTROTYPE   EDITION. 


VAN  ANTWERP,   BRAGG  &  CO., 

CINCINNATI:  NEW    YORK 

137  Walnut  St.  28  Bond  St. 


ECLECTIC  EDUCATIONAL  SERIES, 

Ray's  Mathematical  Series. 

Embracing  a  Thorough,  Progressive,  and  Complete  Course 
in  Arithmetic,  Algebra,  and  the  Higher  Mathematics. 


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FJay's  Series: 

Ray's  New  Primary  Arithmetic 

Ray's  New  Intellectual  Arithmetic 

Ray's  New  Elementary  Arithmetic 

Ray's  New  Practical  Arithmetic 

Key  to  New  Intellectual  and  Practical  Arithmetics 

Ray's  New  Higher  Arithmetic 

Key  to  New  Higher  Arithmetic 

Ray's  New  Test  Examples  in  Arithmetic 

Ray's  Arithmetical  Tablets  (single,  ioc.)  per  doz. . 

Ray's  New  Elementary  Algebra 

Ray's  New  Higher  Algebra    

Key  to  Ray's  New  Algebras   

Ray's  Test  Problems  in  Algebra 

Ray's  Plane  and  Solid  Geometry  {by  Tappan). . . . 
Ray's  Geometry  and  Trigonometry  (by  Tappan) . . 

Ray's  Analytic  Geometry  (by  Howison) 

Ray's  New  Astronomy  (by  Peabody) 

Ray's  Surveying  and  Navigation  (by  Schuyler) .... 
Ray's  Calculus  (by  Clark) 


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Entered  according  to  Act  of  Congress,  in  the  year  1866,  by  Sargent,  Wilson 

&  Hinkle,  in  the  Clerk's  Office  of  the  District  Court  of  the  United 

States  for  the  Southern  District  of  Ohio. 


PREFACE. 


The  object  of  the  study  of  Mathematics  is  two  fold—the  acqui- 
sition of  useful  knowledge,  and  the  cultivation  and  discipline  of 
the  mental  powers.  A  parent  often  inquires,  "Why  should  my 
son  study  Mathematics?  I  do  not  expect  him  to  be  a  surveyor,  an 
engineer,  or  an  astronomer."  Yet,  the  parent  is  very  desirous 
that  his  son  should  be  able  to  reason  correctly,  and  to  exercise, 
in  all  his  relations  in  life,  the  energies  of  a  cultivated  and  disci- 
plined mind.  This  is,  indeed,  of  more  value  than  the  mere  attain- 
ment of  any  branch  of  knowledge. 

The  science  of  Algebra,  properly  taught,  stands  among  the  first 
of  those  studies  essential  to  both  the  great  objects  of  education. 
In  a  course  of  instruction  properly  arranged,  it  naturally  follows 
Arithmetic,  and  should  be  taught  immediately  after  it 

In  the  following  work,  the  object  has  been  to  furnish  an  ele- 
mentary treatise,  commencing  with  the  first  principles,  and  leading 
the  pupil,  by  gradual  and  easy  steps,  to  a  knowledge  of  the  ele- 
ments of  the  science.  The  design  has  been,  to  present  these  in  a 
brief,  clear,  and  scientific  manner,  so  that  the  pupil  should  not  be 
taught  merely  to  perform  a  certain  routine  of  exercises  mechanic- 
ally, but  to  understand  the  why  and  the  wherefore  of  every  step. 
For  this  purpose,  every  rule  is  demonstrated,  and  every  principle 
analyzed,  in  order  that  the  mind  of  the  pupil  may  be  disciplined 
and  strengthened  so  as  to  prepare  him,  either  for  pursuing  the 
study  of  Mathematics  intelligently,  or  more  successfully  attending 
to  any  pursuit  in  life. 

Some  teachers  may  object,  that  this  work  is  too  simple,  and  too 
easily  understood.  A  leading  object  has  been,  to  mSke  the  pupil 
feel,  that  he  is  not  operating  on  unmeaning  symbols,  by  means  of 
arbitrary  rules;  that  Algebra  is  both  a  rational  and  a  practical 
subject,  and  that  he  can  rely  upon  his  reasoning,  and  the  results 

(Hi) 


183990 


iv  PREFACE. 

of  his  operations,  with  the  same  confidence  as  in  arithmetic.  For 
this  purpose,  he  is  furnished,  at  almost  every  step,  with  the  means 
of  testing  the  accuracy  of  the  principles  on  which  the  rules  are 
founded,  and  of  the  results  which  they  produce. 

Throughout  the  work,  the  aim  has  been  to  combine  the  clear 
explanatory  methods  of  the  French  mathematicians  with  the  prac- 
tical exercises  of  the  English  and  German,  so  that  the  pupil  should 
acquire  both  a  practical  and  theoretical  knowledge  of  the  subject 

While  every  page  is  the  result  of  the  author's  own  reflection, 
and  the  experience  of  many  years  in  the  school-room,  it  is  also 
proper  to  state,  that  a  large  number  of  the  best  treatises  on  the 
same  subject,  both  English  and  French,  have  been  carefully  con- 
sulted, so  that  the  present  work  might  embrace  the  modern  and 
most  approved  methods  of  treating  the  various  subjects  presented. 

With  these  remarks,  the  work  is  submitted  to  the  judgment  of 
fellow  laborers  in  the  field  of  education. 

Woodward  College,  August,  1848. 

In  this  New  Electrotype  Edition,  the  whole  volume  has 
been  subjected  to  a  careful  and  thorough  revision.  The  oral 
problems,  at  the  beginning,  have  been  omitted;  the  number  of 
examples  reduced,  where  they  were  thought  to  be  needlessly 
multiplied;  the  rules  and  demonstrations  abridged;  other  methods 
of  proof,  in  a  few  instances,  substituted;  and  questions  for  Gen- 
eral Eeview  introduced  at  intervals,  and  at  the  conclusion.  It 
is  confidently  believed  that  these  modifications,  while  they  do  not 
impair  the  integrity  or  change  the  essential  features  of  the  book, 
will  materially  enhance  its  value,  and  secure  the  approbation 
of  all  intelligent  teachers. 

March,  1866. 

To  Teachers. — The  following  subjects  may  be  omitted  by  the  younger  pupils,  and 
passed  over  by  those  more  advanced,  until  the  book  is  reviewed :  Observations  on 
Addition  and  Subtraction,  Articles  60 — 64;  the  greater  part  of  Chapter  II. ;  supple- 
ment to  Simple  Equations,  Articles  164 — 177  ;  properties  of  the  Roots  of  an  Equa- 
tion of  the  Second  Degree,  Articles  215 — 217. 

The  pupil  should  be  exercised  in  the  solution  of  examples,  until  the  principles 
are  thoroughly  understood ;  and,  in  the  review,  he  should  be  required  to  demon- 
strate the  rules  on  the  blackboard. 


CONTENTS 


I.— FUNDAMENTAL  KULES.  page. 

Definitions 7 

Explanation  of  Signs  and  Terms 9 

Addition 17 

Subtraction 22 

Observations  on  Addition  and  Subtraction 27 

Multiplication 30 

Division 39 

II.— THEOREMS,  FACTORING,  Etc. 

Algebraic  Theorems 46 

Factoring 50 

Greatest  Common  Divisor , 55 

Least  Common  Multiple 62 

III.— ALGEBRAIC  FRACTIONS. 

Definitions  and  Fundamental  Propositions 65 

To  reduce  a  Fraction  to  its  Lowest  Terms 69 

To  reduce  a  Fraction  to  an  Entire  or  Mixed  Quantity 70 

To  reduce  a  Mixed  Quantity  to  a  Fraction 71 

Signs  of  Fractions 72 

To  reduce  Fractions  to  a  Common  Denominator 74 

To  reduce  Fractions  to  the  Least  Common  Denominator  ....  76 

To  reduce  a  Quantity  to  a  Fraction  with  a  given  Denominator      .  77 

To  convert  a  Fraction  to  another  with  a  given  Denominator    .     .  77 

Addition  and  Subtraction  of  Fractions 78 

Multiplication  of  Fractions 81 

Division  of  Fractions       84 

To  reduce  a  Complex  Fraction  to  a  Simple  one 87 

Resolution  of  Fractions  into  Series 88 

IV.— SIMPLE  EQUATIONS. 

Definitions  and  Elementary  Principles 90 

Transposition 93 

To  clear  an  Equation  of  Fractions 94 

Simple  Equations  of  one  Unknown  Quantity 96 

Simple  Equations  of  two  Unknown  Quantities 114 

Elimination  by  Substitution 114 

Elimination  by  Comparison 115 

(v) 


vi  CONTENTS. 

PAGE. 

Elimination  by  Addition  and  Subtraction 117 

Problems  producing  Equations  of  two  Unknown  Quantities     .     .  118 

Equations  containing  three  or  more  Unknown  Quantities  .     .     .  126 

V.— SUPPLEMENT  TO  SIMPLE  EQUATIONS. 

Generalization 135 

Negative  Solutions 145 

Discussion  of  Problems 146 

Problem  of  the  Couriers 14., 

Cases  of  Indetermination  and  Impossible  Problems 151 

VI.— OF   POWERS,  ROOTS,  AND  RADICALS. 

Involution  or  Formation  of  Powers 154 

To  raise  a  Monomial  to  any  given  Power 154 

To  raise  a  Polynomial  to  any  given  Power 156 

To  raise  a  Fraction  to  any  Power 157 

Binomial  Theorem 157 

Evolution 163 

Square  Root  of  Numbers 163 

Square  Root  of  Fractions 167 

Perfect  and  Imperfect  Squares 168 

Approximate  Square  Roots 169 

Square  Root  of  Monomials 172 

Square  Root  of  Polynomials 173 

Radicals  of  the  Second  Degree 177 

Reduction  of  Radicals 178 

Addition  of  Radicals 180 

Subtraction  of  Radicals 181 

Multiplication  of  Radicals 182 

Division  of  Radicals 184 

Simple  Equations  containing  Radicals  of  the  Second  Degree  .     .  186 

VIL— QUADRATIC  EQUATIONS. 

Definitions  and  Elementary  Principles 189 

Pure  Quadratics 190 

Affected  Quadratics *.     ...  194 

Hindoo  Method  of  solving  Equations  of  the  Second  Degree     .     .  199 

Properties  of  the  Roots  of  an  Affected  Quadratic 204 

Quadratic  Equations  containing  two  Unknown  Quantities  .     .     .  210 

VIII.— PROGRESSIONS  AND  PROPORTION. 

Arithmetical  Progression 216 

Geometrical  Progression 223 

Ratio 228 

Proportion 231 


^     OF  THE 

UNIVERSITY 

OF 

ILIFC 


ELEMENTS  OF  ALGEBRA. 


I.  DEFINITIONS. 

Note  to  Teachers. — Articles  1  to  15  may  be  omitted  until  the 
pupil  reviews  the  book. 

Article  1.  In  Algebra,  quantities  are  represented  by 
letters  of  the  alphabet. 

2.  Quantity  is  any  thing  that  is  capable  of  increase  or 
decrease ;  as,  numbers,  lines,  space,  time,  etc. 

3.  Quantity  is  called  magnitude,  when  considered  in  an 
undivided  form;  as,  a  quantity  of  water. 

4.  Quantity  is  called  multitude,  when  made  up  of  in- 
dividual and  distinct  parts ;  as,  three  cents,  a  quantity 
composed  of  three  single  cents. 

5.  One  of  the  single  parts  of  which  a  quantity  of  multi- 
tude is  composed,  is  called  the  unit  of  measure;  thus,  1  cent 
is  the  unit  of  measure  of  the  quantity  3  cents. 

The  value  or  measure  of  any  quantity  is  the  number  of 
times  it  contains  its  unit  of  measure. 

6.  In  quantities  of  magnitude,  where  there  is  no  nat- 
ural unit,  it  is  necessary  to  fix  upon  an  artificial  unit  as 
a  standard  of  measure ;  then,  to  find  the  value  of  the 
quantity,  we  ascertain  how  many  times  it  contains  its  unit 
of  measure.     Thus, 

To  measure  the  length  of  a  line,  take  a  certain  assumed 

Review. — 1.  How  are  quantities  represented  in  Algebra  ?  2.  What 
is  quantity?     3.  When  called  magnitude? 

4.  When  multitude?  5.  What  is  the  unit  of  measure?  6.  How 
find  the  value  of  a  quantity  when  there  is  no  natural  unit? 

7 


8  RAY'S  ALGEBRA,  FIRST  BOOK. 

distance  called  a  foot,  and,  applying  it  a  certain  number 
of  times,  say  5,  it  is  found  that  the  line  is  5  feet  long ;  in 
this  case,  1  foot  is  the  unit  of  measure. 

7.  The  Numerical  Value  of  a  quantity  is  the  num- 
ber that  shows  how  many  times  it  contains  its  unit  of 
measure. 

Thus,  the  numerical  value  of  a  line  5  feet  long,  is  5. 
The  same  quantity  may  have  different  numerical  values, 
according  to  the  unit  of  measure  assumed. 

8.  A  Unit  is  a  single  thing  of  an  order  or  kind. 

0.  Number  is  an  expression  denoting  a  unit,  or  a  collec- 
tion of  units.     Numbers  are  either  abstract  or  concrete. 

10.  An  Abstract  Number  denotes  how  many  times  a 
unit  is  to  be  taken. 

A  Concrete  Number  denotes  the  units  that  are  taken. 

Thus,  4  is  an  abstract  number,  denoting  merely  the  num- 
ber of  units  taken;  while  4  feet  is  a  concrete  number,  denot- 
ing what  unit  is  taken,  as  well  as  the  number  taken. 

Or,  a  concrete  number  is  the  product  of  the  unit  of  measure 
by  the  corresponding  abstract  number.  Thus,  $6  equal  $1 
multiplied  by  6,  or  $1  taken  6  times. 

11.  In  algebraic  computations,  letters  are  considered 
the  representatives  of  numbers. 

12.  There  are  two  kinds  of  questions  in  Algebra,  the- 
orems and  problems. 

13.  In  a  Theorem,  it  is  required  to  demonstrate  some 
relation  or  property  of  numbers,  or  abstract  quantities. 

14.  In  a  Problem,  it  is  required  to  find  the  value  of 
some  unknown  quantity,  by  means  of  certain  given  rela- 
tions existing  between  it  and  others,  which  are  known. 

Review. — 7.  Define  numerical  value.  8.  What  is  a  unit?  9.  A 
number?  10.  What  does  an  abstract  number  denote?  A  concrete 
number?  11.  What  do  the  letters  used  in  Algebra  represent? 
12.  How  many  kinds  of  questions  in  Algebra?  13.  What  is  a  the- 
orem?    14.  A  problem? 


DEFINITIONS  AND  NOTATION.  0 

1*5.  Algebra  is  a  general  method  of  solving  problems 
and  demonstrating  theorems,  by  means  of  figures,  letters, 
and  signs.     The  letters  and  signs  are  called  symbols. 

EXPLANATION  OF  SIGNS  AND  TERMS. 

1G.  Known  Quantities  are  those  whose  values  are 
given ;  Unknown  Quantities,  those  whose  values  are  to  be 
determined. 

17.  Known  quantities  are  generally  represented  by  the 
first  letters  of  the  alphabet,  as  a,  b,  c,  etc. ;  unknown  quan- 
tities, by  the  last  letters,  as  x,  y,  z. 

18.  The  principal  signs  used  in  Algebra  are 

— »     ~r?         ?      Xi     -s->     C   ji     ^j      v ' 

Each  sign  is  the  representative  of  certain  words.  They 
are  used  to  express  the  various  operations  in  the  clearest 
and  briefest  manner. 

19.  The  Sign  of  Equality,  =,  is  read  equal  to.  It  de- 
notes that  the  quantities  between  which  it  is  placed  are 
equal.  Thus,  a=3,  denotes  that  the  quantity  represented 
by  a  is  equal  to  3. 

20.  The  Sign  of  Addition,  -f ,  is  read  plus.  It  denotes 
that  the  quantity  to  which  it  is  prefixed  is  to  be  added. 

Thus,  a-\-b,  denotes  that  b  is  to  be  added  to  a.  If 
a=2  and  b=S,  then  a-j-6=2-j-3,  which  =5. 

21.  The  Sign  of  Subtraction,  — ,  is  read  minus.  It  de- 
notes that  the  quantity  to  which  it  is  prefixed  is  to  be 
subtracted. 

Thus,  a — Z>,  denotes  that  b  is  to  be  subtracted  from  a. 
If  a=5  and  Z>=3,  then  5  —  3=2. 

Review.— 15.  What  is  Algebra?  What  are  symbols ?  lG.Wh.it 
are  known  quantities?  Unknown  quantities?  17.  By  what  are 
known  quantities  represented?     Unknown  quantities? 

18.  Write  the  principal  signs  used  in  Algebra.  What  does  each 
represent?     For  what  used? 

19.  How  is  the  sign  =  read?  What  does  it  denote?  20.  How  is 
the  sign  -f-  read?  What  denote?  21.  How  is  the  sign  —  read? 
What  denote? 


10  RAYS  ALGEBRA,  FIRST  BOOK. 

22.  The  signs  -f-  and  —  are  called  the  signs.  The 
former  is  called  the  positive,  the  latter  the  negative  sign : 
they  are  said  to  be  contrary  or  opposite. 

23.  Every  quantity  is  supposed  to  be  preceded  by  one 
of  these  signs.  Quantities  having  the  positive  sign  are 
called  positive ;  those  having  the  negative  sign,  negative. 

When  a  quantity  has  no  sign  prefixed,  it  is  positive. 

24.  Quantities  having  the  same  sign  are  said  to  have 
lilze  signs ;  those  having  different  signs,  unlike  signs. 

Thus,  -\-a  and  -{-&,  or  — a  and  — h,  have  like  signs  ; 
■while  -j-c  and  — d  have  unlike  signs. 

25.  The  Sign  of  Multiplication,  X ,  is  read  into,  or  mul- 
tiplied by.  It  denotes  that  the  quantities  between  which 
it  is  placed  are  to  be  multiplied  together. 

The  product  of  two  or  more  letters  is  sometimes  ex- 
pressed by  a  dot  or  point,  but  more  frequently  by  writing 
them  in  close  succession  without  any  sign.  Thus,  ah  ex- 
presses the  same  as  «X&  ov  a.b,  and  abc=ayiby^c,  or  a,.b.c. 

26.  Factors  are  quantities  that  are  to  be  multiplied 
together. 

The  continued  product  of  several  factors  means  the  prod- 
uct of  the  first  and  second  multiplied  by  the  third,  this 
product  by  the  fourth,  and  so  on. 

Thus,  the  continued  product  of  a,  b,  and  c,  is  «X&X^> 
or  abc.  If  a=2,  6=3,  and  c=5,  then  a&c=2x3x5= 
6x5=30. 

27.  The  Sign  of  Division,  -s-,  is  read  divided  by.     It 

Review. — 22.  What  are  the  signs  plus  and  minus  called',  by  way 
of  distinction?     Which  is  positive?     Which  negative? 

23.  What  are  quantities  preceded  by  the  sign  plus  said  to  be?  By 
the  sign  minus?  When  no  sign  is  prefixed?  24.  When  do  quan- 
tities have  like  signs?     When  unlike  signs? 

25.  How  is  the  sign  X  read,  and  what  does  it  denote?  What  other 
methods  of  representing  multiplication?  26.  What  are  factors? 
How  many  in  a?     In  ab?     In  abc?     In  babe? 

27.  How  is  the  sign  -f-  read,  and  what  does  it  denote  ?  What 
other  methods  of  representing  division? 


DEFINITIONS  AND  NOTATION.  11 

denotes  that  the  quantity  preceding  it  is  to  be  divided  03/ 
that  following  it.  Division  is  oftener  represented  by  plac- 
ing the  dividend  as  the  numerator,  and  the  divisor  as  the 
denominator  of  a  fraction. 

Thus,  «-=-&,  or  7,  means,  that  a  is  to  be  divided  by  b. 
Ir*a=I2  and  Z>=3,  then  a~b=12~S=4: ;  or*=M-L±. 

Division  is  also  represented  thus,  o]&,  or  b\ay  a  denoting 
the  dividend,  and  b  the  divisor. 

28.  The  Sign  of  Inequality,  >,  denotes  that  one  of 
the  two  quantities  between  which  it  is  placed  is  greater 
than  the  other.  The  opening  of  the  sign  is  toward  the 
greater  quantity. 

Thus,  a>&,  denotes  that  a  is  greater  than  b.  It  is  read, 
a  greater  than  b.  If  a=b,  and  &— 3,  then  5>3.  Also, 
c<^d,  denotes  that  c  is  less  than  d.  It  is  read,  c  less  than  d. 
If  c=4  and  d=1,  then  4<7. 

29.  The  Sign  of  Infinity,  od,  denotes  a  quantity  greater 
than  any  that  can  be  assigned,  or  one  indefinitely  great. 

30.  The  Numeral  Coefficient  of  a  quantity  is  a  num- 
ber prefixed  to  it,  showing  how  many  times  the  quantity 
is  taken. 

Thus,  a-\-a-\-a-\-a—Aa  ;  and  ax-\-ax-\-ax=^2>ax. 

31.  The  Literal  Coefficient  of  a  quantity  is  a  quantity 
by  which  it  is  multiplied.  Thus,  in  the  quantity  ay,  a  may 
be  considered  the  coefficient  of  y,  or  y  the  coefficient  of  a. 

The  literal  coefficient  is  generally  regarded  as  a  known 
quantity. 

32.  The  coefficient  of  a  quantity  may  consist  of  a  num- 
ber  and   a    literal   part.     Thus,   in    &ax,   5a  may  be   re- 

Review. — 28.  What  is  the  sign  >  called,  and  what  does  it  de- 
note?    Which  quantity  is  placed  at  the  opening? 

29.  What  does  the  sign  qd  denote?  30.  What  is  a  numeral  co- 
efficient?    How  often  is  ax  taken  in  Sax?     In  5ax?     In  lax? 

31.  What  is  a  literal  coefficient?  32.  When  a  quantity  has  no 
coefficient,  what  is  understood? 


12  RAYS  ALGEBRA,  FIRST  BOOK. 

garded  as  the    coefficient   of  x.      If  a=2,  then   5a=10, 
and  5ax=10x. 

When  no  numeral  coefficient  is  prefixed  to  a  quantity, 

its  coefficient   is  understood  to    be   unity.  Thus,  a=la, 

and  bx=lbx. 

■\ 

33.  The  Power  of  a  quantity  is  the  product  arising  from 
multiplying  the  quantity  by  itself  one  or  more  times. 

When  the  quantity  is  taken  twice  as  a  factor,  the  prod- 
uct is  called  its  square,  or  second  power ;  when  three  times, 
the  cube,  or  third  power ;  when  four  times,  the  fourth 
power,  and  so  on. 

Thus,  ay^a=aa,  is  the  second  power  of  a ;  aX«Xa= 
aaa,  is  the  third  power  of  a;  ay^ay^aY^a=aaaa,  is  the 
fourth  power  of  a. 

An  Exponent  is  a  figure  placed  at  the  right,  and  a  little 
above  a  quantity,  to  show  how  many  times  it  is  taken  as 
a  factor. 

Thus,  aa=a?',  aaa=a3;  aaaa=a*;  aabbb=aQb2. 

When  no  exponent  is  expressed,  it  is  understood  to  be 
unity.  Thus,  a  is  the  same  as  a1,  each  expressing  the  first 
power  of  a. 

34.  To  raise  a  quantity  to  any  given  power  is  to  find 
that  power  of  the  quantity. 

35.  The  Root  of  a  quantity  is  another  quantity,  some 
power  of  which  equals  the  given  quantity.  The  root  is 
called  the  square  root,  cube  root,  fourth  root,  etc.,  accord- 
ing to  the  number  of  times  it  is  taken  as  a  factor  to  pro- 
duce the  given  quantity. 

Thus,  a  is  the  second  or  square  root  of  a2,  since  ay(a=a2. 
So,  x  is  the  third  or  cube  root  of  #3,  since  xX%~Xx=x3. 

36.  To  extract  any  root  of  a  quantity  is  to  find  that  root. 

Eeview. — 33.  What  is  the  power  of  a  quantity?  What  is  the 
second  power  of  a?     The  third  power  of  a \ 

33.  What  is  an  exponent?  For  what  used?  How  many  times 
is  x  taken  as  a  factor  in  x2  ?  In  a;3?  In  x5  ?  Where  no  exponent 
is  written,  what  is  understood  ?     35.  What  is  the  root  of  a  quantity  ? 


DEFINITIONS  AND  NOTATION.  13 

37.  The  Radical.  Sign,  -j/,  placed  before  a  quantity,  in- 
dicates that  its  root  is  to  be  extracted. 

Thus,  pKa,  or  |/a,  denotes  the  square  root  of  a;  ^Ka,  de- 
notes the  cube  root  of  a;  -j^a,  denotes  the  fourth  root  of  a. 

38.  The  number  placed  over  the  radical  sign  is  called 
the  index  of  the  root.  Thus,  2  is  the  index  of  the  square 
root,  3  of  the  cube  root,  4  of  the  fourth  root,  and  so  on. 
When  the  radical  has  no  index  over  it,  2  is  understood. 

30.  Every  quantity  or  combination  of  quantities  ex- 
pressed by  means  of  symbols,  is  called  an  algebraic  ex- 
pression. 

Thus,  3a  is  the  algebraic  expression  for  3  times  the 
quantity  a;  3a — 46,  for  3  times  a,  diminished  by  4  times  b; 
2a2-\-3ab,  for  twice  the  square  of  a,  increased  by  3  times 
the  product  of  a  and  b. 

4©.  A  Monomial,  or  Term,  is  an  algebraic  expression, 
not  united  to  any  other  by  the  sign  -f-  or  — . 

A  monomial  is  sometimes  called  a  simple  quantity.  Thus, 
a,  3a,  — a2b,  2any2,  are  monomials,  or  simple  quantities. 

41.  A  Polynomial  is  an  algebraic  expression,  composed 
of  two  or  more  terms. 

Thus,  c-\-2d — b  is  a  polynomial. 

42.  A  Binomial  is  a  polynomial  composed  of  two  terms. 
Thus,  a-\-b,  a — b,  and  c2 — d,  are  binomials. 

A  Residual  Quantity  is  a  binomial,  in  which  the  second 
term  is  negative,  as  a — b. 

43.  A  Trinomial  is  a  polynomial  consisting  of  three 
terms.     Thus,  a-j-6-j-c,  and  a — b — c,  are  trinomials. 

44.  The  Numerical  Value  of  an  algebraic  expression 

Review. — 37.  What  is  the  sign  j/  called,  and  what  does  it  de- 
note? 38.  What  is  the  number  placed  over  the  radical  sign  called? 
39.  What  is  an  algebraic  quantity?  40.  A  monomial?  A  simple  quan- 
tity?   41.  A  polynomial?    42.  A  binomial?    A  residual  quantity? 

43.  A  trinomial?  44.  What  is  the  numerical  value  of  an  alge- 
braic expression? 


14  RAY'S  ALGEBRA,  FIRST  BOOK. 

is  the  number  obtained,  by  giving  particular  values  to  the 
letters,  and  then  performing  the  operations  indicated. 

In  the  algebraic  expression  2a-j-36,  if  a=4,  and  Z>=5, 
then  2a=8,  and  36=15,  and  the  numerical  value  is 
8+15=23. 

45.  The  value  of  a  polynomial  is  not  affected  by  chang- 
ing the  order  of  the  terms,  provided  each  term  retains  its 
respective  sign.  Thus,  a2-\-2a-\-b=b-\-a2-\-2a.  This  is 
self-evident. 

46.  Each  of  the  literal  factors  of  any  simple  quantity 
or  term  is  called  a  dimension  of  that  term.  The  degree  of 
a  term  depends  on  the  number  of  its  literal  factors. 

Thus,  ax  consists  of  two  literal  factors,  a  and  x,  and  is 
of  the  second  degree.  The  quantity  a?b  contains  three  lit- 
eral factors,  a,  a,  and  b,  and  is  of  the  third  degree.  2a?x^ 
contains  5  literal  factors,  a,  a,  a,  a;,  and  x,  and  is  of  the 
fifth  degree;  and  so  on. 

47.  A  polynomial  is  said  to  be  homogeneous,  when  each 
of  its  terms  is  of  the  same  degree. 

Thus,  the  polynomials  2a — 3Z>-f-c,  of  the  first  degree, 
a*-j-3bc-{-xy,  of  the  second  degree,  and  xz — 8a?/2,  of  the 
third  degree,  are  homogeneous  :  a3-f  re2  is  not  homogeneous. 

48.  A  Parenthesis,  (  ),  is  used  to  show  that  all  the 
included  terms  are  to  be  considered  together  as  a  single 
term. 

Thus,  4  (a — b)  means  that  a — b  is  to  be  multiplied  by  4; 
(a-J-#)  (a — x)  means  that  a-\-x  is  to  be  multiplied  by 
a — x;  10 — («-|-c)  means  that  a-\-c  is  to  be  subtracted 
from  10;  (a — b)2  means  that  a — b  is  to  be  raised  to  the 
second  power;  and  so  on. 

49.  A  Vinculum,  ,  is  sometimes  used  instead  of 

Review. — 46.  What  is  the  dimension  of  a  term?  On  what  docs 
the  degree  of  a  term  depend?  What  is  the  degree  of  the  term  zy? 
Of  zyz?  Of  2axy?  Of  x2?  47.  When  is  a  polynomial  homo- 
geneous? 48.  For  what  is  a  parenthesis  used?  40.  What  i&.  d  via' 
culum,  and  for  what  used  ? 


DEFINITIONS  AND  NOTATION.  15 


a  parenthesis.  Thus,  a — by^x  means  the  same  as  (a — h)x. 
Sometimes  the  vinculum  is  placed  vertically:  it  is  then 
called  a  bar. 

Tims,     a  if  has  the  same  meaning  as  (a — a-f-4)/2. 

— x 

+  4_ 

50.  Similar  or  Like  quantities  are  those  composed  of 
the  same  letters,  affected  with  the  same  exponents. 

Thus,  'lab  and  — Sab,  also  4a3&2  and  7a3b'\  are  similar 
terms;  but  2a2b  and  2ab2  are  not  similar;  for,  though 
composed  of  the  same  letters,  these  letters  have  different 
exponents. 

51.  The  Reciprocal  of  a  quantity  is  unity  divided  by  that 

quantity.     Thus,  the  reciprocal  of  2  is  3,  of  a  is  -. 

The  reciprocal  of  |  is  1  divided  by  § ,  or  1.  Hence, 
the  reciprocal  of  a  fraction  is  the  fraction  inverted. 

52.  The  same  letter  accented  is  often  used  to  denote 
quantities  which  occupy  similar  positions  in  different  equa- 
tions or  investigations. 

Thus,  a,  a/  a,"  a'",  represent  four  different  quantities ; 
read  a,  a  prime,  a  second,  a  third,  and  so  on. 

EXAMPLES. 

The  following  examples  are  intended  to  exercise  the 
learner  in  the  use  and  meaning  of  the  signs. 

Copy  each  example  on  the  slate  or  blackboard,  and  then 
express  it  in  common  language. 

Let  the  numerical  values  be  found,  on  the  supposition 
that  a=4,  Z>=3,  c=5,  d=10,  x=2,  and  y=6. 

1.  c+d—b.    .   .  Ans.  12. 

2.  4a—  x.    .   .   .  Ans.  14. 


3.  —  Sax.    .  .Ans.  —24. 

4.  6a2x     .   .   .  Ans.  192. 


5.  ^+^ Ans.  33. 

b         x 

6.  3a2+2cx—b*  .  Ans.  41. 

7.  «(«+&) Ans.  28. 


Review. — 50.  What  are  similar  or  like  quantities?     51.  The  re- 
ciprocal of  a  quantity?     52.  What  the  use  of  accented  letters? 


16  RAYS  ALGEBRA,  FIRST  BOOK. 

8.  a+iXa-^ Ans-  13- 

9.  (a+6)(a— &) Ans.  1. 

10.  x2— 3(a-\-x)(a— x)-\-2by Ans.  4. 

11.  - — 6xy/a Ans.  — 16. 

(a — xy  v 

12.  3(a-f-c)(«— c)-f-3a2— 3c2 Ans.  —54. 

13.  ,     4-a—x Ans.  4. 

a-\-x 

In  the  following,  convert  the  words  into  algebraic  symbols : 

1.  Three  times  a,  plus  b,  minus  four  times  c. 

2.  Five  times  a,  divided  by  three  times  b. 

3.  a  minus  b,  into  three  times  c. 

4.  a,  minus  three  times  b  into  c. 

5.  a  plus  &,  divided  by  three  c. 

6.  a,  plus  b  divided  by  three  c. 

7.  a  squared,  minus  three  a  into  b,  plus  5  times  c  into 
c?  squared. 

8.  x  cubed  minus  b  cubed,  divided  by  x  squared  minus 
b  squared. 

9.  Five  a  squared,  into  a  plus  b,  into  c  minus  d,  minus 
three  times  x  fourth  power. 

10.  a   squared   plus  b   squared,    divided   by  a   plus    6, 
squared. 

11.  The  square  root  of  a,  minus  the  square  root  of  x. 

12.  The  square  root  of  a  minus  x. 

ANSWERS. 

1.  ga+fc— 4c.  7.  a2— Sab+bccP. 

9    5a  ft     <*— h* 

*'  37/  *•  ~tf-U>' 

3.  (a— &)3r.  9.  5a2(«-f-Z>)  (c— d)— SxK 

4.  a— 36c.  a2+&2 


&    «+&  •  (a+&)2 


3c  '  11.  j/a — -y/x. 

b 
8c 


6.a  +  i  12V(a-*). 


ADDITION.  17 


ADDITION. 

53.  Addition,  in  Algebra,  is  the  process  of  finding 
the  simplest  expression  for  the  sum  of  two  or  more  alge- 
braic quantities. 

CASE    I. 

When  the  Quantities  are  similar,  and  have  the  same  Sign. 

1.  James  has  3  pockets,  each  containing  apples :  in  the 
first  he  has  3  apples,  in  the  second  4,  and  in  third  5. 

In  order  to  find  how  many  apples  he  has,  suppose  he  proceeds  to 
find  their  sum  in  the  following  manner :     3  apples, 

4  apples, 

5  apples, 

12  apples. 
But,  instead  of  writing  the  word  apples,  suppose  he  should  use  the 
letter  a,  thus:  3a 

4a 
5a 

~12a 
It  is  evident  that  the  sum  of  3  times  a,  4  times  a,  and  5  times  a, 
is  12  times  a,  or  12a,  whatever  a  may  represent. 

2.  In  the  same  manner  the  sum  of  — 3a,  — 4a,  — 3a 
and  — 5a  would  be  — 12a.     Hence,  — 4a 

— 5a 


-12a 

TO  ADD  SIMILAR  QUANTITIES  WITH  LIKE  SIGNS, 

Rule. — Add  together  the  coefficients  of  the  several  quan- 
tities; to  their  sum  prefix  the  common  sign,  and  annex  the 
common  letter  or  letters. 

Note. — When  a  quantity  has  no  coefficient,  1  is  understood; 
thus,  a=la. 

Review. — 53.  What  is  algebraic  addition?  When  quantities  are 
similar  and  have  the  same  sign,  how  are  they  added  together  ? 

When  several  quantities  are  to  be  added  together,  is  the  result 
affected  by  the  order  in  which  they  are  taken  ? 
1st  Bk.  2 


18  RAY'S  ALGEBRA,  FIRST  BOOK. 


EXAMPLES. 

(3) 

(4) 

(5) 

(6) 

3a 

— 6xy 

2a2 

—  3a2Z> 

2a 

—xy 

3a2 

— 4a2Z> 

a 

—4xy 

5a2 

— 5a2Z> 

ba 

-Sxy 

7a2 

—  2a*b 

Sum=lla  —  14xy  17a2  — 14a26 

In  the  third  example,  suppose  a=2,  then  3a=3x2=6,  2a=2x 
2=4,  a=2,  5a=5x2=10;  their  sum  is  6+4-j-2-j-10=22. 

But  the  sum,  22,  is  more  easily  found  from  the  algebraic  sum,  11a, 
forlla=llx2=22. 

In  the  fourth  example,  let  #=3  and  ?/=2;  then, 
— 6xy=— 6  X  3  X  2=— 36 
_  xy=—      3X2=-  6 
— 4z?/=— 4  X  3  X  2=— 24 
_3:ry =— 3  X3  X  2=— 1 8 

the  sum  of  their  values        = — 84. 

But  this  is  more  easily  found  thus:  — 14x?/= — 14x3X2= — 84. 
In  the  fifth  example,  let  a  represent  three  feet;   then, 

2a2=2aa=2x3x3=18  square  feet, 

3a2=3aa=3x3x3=27      «         " 

5a2=5aa=5x  3x3=45      "         " 

7a2=7aa=7x3x3=63      «         « 

and  their  sum  is  153       u  a 

Or  the  sum  =l7a2=l7x3x3=153  square  feet, 

Note. — Let  the  learner  test  the  following  examples  numerically, 
by  assigning  values  to  the  letters. 

7.  What  is  the  sum  of  36,  5Z>,  lb,  and  9b?     Ans.  247;. 

8.  Of  2ab,  bab,  Sab,  and  lla&?  Ans.  26ab. 

9.  Of  abc,  dabc,  ^abc,  and  12aZ>c?  Ans.  2Sabc. 
10.  Of  —  by,  —2by,  —My,  and  —  %by?     Ans.  —  16by. 

(11)  (12)  (13) 

3ay-f7  Sx—4y  Sa2—2ax 

oy-j-8  bx — By  5a2 — Sax 

2ay-f-4  *?x— Gy  7a2— hax 

5a_y-j-6  6x — 2y  4a2 — 4*KB 


ADDITION.  10 

CASE    II. 

54.   Wlien  Quantities  are  alike,  but  have  Unlike  Signs. 

1.  James  receives  from  one  man  6  cents,  from  another 
9,  and  from  a  third  10.  He  spends  4  cents  for  candy 
and  3  for  apples :  how  much  will  he  have  left  ? 

If  the  quantities  lie  received  be  considered  positive,  those  he  spent 
may  be  considered  negative;  and  the  question  is,  to  find  the  sum 
of  -{~6c,  -r~9<?,  -f-10c,  — 4c  and  — 3c,  which  may  be  written  thus : 

+6c 
,g  It  is  evident  the  true  result  will  be  found  by 

4-1  (V      collecting  the  positive  quantities  into  one  sum, 

a        and   the    negative   quantities  into   another,    and 

o        taking  their  difference.     It  is  thus  found  that  he 

received  25c,  and  spent  7c,  which  left  18c. 


-{-18c 

2.  Suppose  James  should  receive  5  cents,  and  spend 
*I  cents,  what  sum  would  he  have  left  ? 

If  we  denote  the  5c  as  positive,  the  7c  will  be  negative,  and  it  is 
required  to  find  the  sum  of  -j-oc  and  — 7c. 

In  its  present  form  it  is  evident  that  the  question  is  impossible. 
But  if  we  suppose  that  James  had  a  certain  sum  of  money  before 
he  received  the  5c,  we  may  inquire  what  effect  the  operation  had 
upon  his  money. 

The  answer  obviously  is,  that  his  money  was  diminished  2  cents; 
this  would  be  indicated  by  the  sum  of  -|-5c  and  — 7c,  being  — 2c. 

Hence,  we  say  that  the  sum  of  a  positive  and  negative  quantity 
is  equal  to  the  difference  between  the  two;  the  object  being  to  find 
what  the  united  effect  of  the  two  will  be  upon  some  third  quantity. 

This  may  be  further  illustrated  by  the  following  example : 

3.  A  merchant  has  a  certain  capital;  during  the  year  it 
is  increased  by  3a  and  8a  $'s,  and  diminished  by  2a  and 
5a  $'s  :  how  much  will  it  be  increased  or  diminished  at 
the  close  of  the  year? 

If  we  call  the  gains  positive,  the  losses  will  be  negative.  The  sum 
of  +3(2,  -f-8a,  —2a,  and  —5a,  is  -j-lla— 7a=4r4a. 

Hence,  we  say  that  the  merchant's  capital  will  be  increased  by  4a 
$'s,  which  is  the  same  as  to  increase  it  by  3a  and  8a  $'s,  and  then 
diminish  it  by  2a  and  5a   $'s. 


23  RAY'S  ALGEBRA,  FIRST  BOOK. 

Had  the  loss  been  greater  than  the  gain,  the  capital  would  have 
been  diminished,  and  the  result  would  have  been  negative. 

If  the  gain  and  loss  were  equal,  the  capital  would  neither  be  in- 
creased nor  diminished,  and  the  sum  of  the  positive  and  negative 
quantities  would  be  0.     Thus,  -\-3ci — 3a=0. 

Hence,  to  add  a  negative  quantity  is  the  same  as  to  subtract  a 
positive  quantity.  In  such  cases,  the  process  is  called  algebraic 
addition,  and  the  sum  the  algebraic  sum,  to  distinguish  them  from 
arithmetical  addition  and  arithmetical  sum.     Hence, 


TO   ADD   LIKE   QUANTITIES  WHICH   HAVE  UNLIKE  SIGNS, 

Rule. — 1.  Find  the  sum  of  the  coefficients  of  the  positive 
quantities;  also,  the  sum  of  the  coefficients  of  the  negative 
quantities. 

2.  Subtract  the  less  sum  from  the  greater;  to  the  difference 
prefix  the  sign  of  the  greater,  and  annex  the  common  literal 
part. 

4.  What  is  the  sum  of  -j-3a,  — 5a,  -(-9a,  — 6a,  and  -f  7a? 

Sum  of  positive  coefficients,  3-j-9-|—7=-}— 1 9. 
Sum  of  negative  coefficients,    — 5 — 6= — 11. 

Difference,  =19 — 11=8.  Prefixing  the  sign  of  the  greater,  and 
annexing  the  literal  part,  we  have  for  the  required  sum  -\-8a. 

In  practice,  it  is  most  convenient  to  write  the  3a 

different  terms  under  each  other.     Thus,  — 5a 

9a 

—6a 

7a 


Sum=8a 


EXAMPLES. 


5.  What  is  the  sum  of  8a  and  — 5a?  Ans.  3a. 

6.  Of  5a  and  —8a?  Ans.  — 3«. 

7.  Of  — 7 ax,  Sax,  6ax,  and  — ax?  Ans.  ax. 

8.  Of  5abx,  — 7abx,  Sabx,  — abx,  and  4afoe?  Ans.  4abx. 

9.  Of  6a— 46,  3a-j-26,  —7a—8b,  and  —  a+9/>? 

Ans.  a — b. 
Review. — 54.  What  is  case  2d  in  Addition?     The  rule? 


ADDITION.  21 

10.  Of  4a2— 26,  —  6a2+26,  2a2— 36,  —5a2— 86,  and 
— 3a2+96?  Ans.  —8a2— 26. 

11.  Of  xy — ac,  3xy — 9ac,  — ^xy-\-hac,  4txy-{-6ac,  and 
— xy — 2ac?  Ans.  — ac. 

Note. — The  operation  of  collecting  an  algebraic  expression  into 
one  sum  is  called  the  Reduction  of  Polynomials.  The  following  are 
examples : 

12.  Reduce  3a6-j-5c— 7a6+8c-f  8a6— 14c— 2a6+c  to 
its  simplest  form.  Ans.  2a6. 

13.  Reduce  5a2c— 362-J-4a2c-f-562—  8a2c+262  to  its  sim- 
plest form.  Ans.  a2c-}-462. 

CASE    III. 

5I>«  When  the  Quantities  are  Unlike,  or  partly  Like  and 
partly  Unlike. 

1.  Thomas  has  a  marbles  in  one  hand,  and  b  marbles 
in  the  other:  what  expression  will  represent  the  number 
in  both? 

If  a  is  represented  by  3,  and  b  by  4,  then  the  number  in  both 
would  be  represented  by  3-}-4,  or  7. 

Or,  the  number  in  both  would  be  represented  by  a-\-b ;  but  unless 
the  numerical  values  of  a  and  6  are  given,  it  is  evidently  impos- 
sible to  represent  their  sum  more  concisely  than  by  a-\-b. 

So,  the  sum  of  a-\-b  and  c-\-d,  is  represented  by  a-\-b-\-c-\-d. 

If,  in  any  expression,  there  are  like  quantities,  it  is  obvious  that 
they  may  be  added  by  the  preceding  rules.  Case  3d,  therefore,  em- 
braces the  two  preceding  cases.     Hence, 

TO   ADD    ALGEBRAIC   QUANTITIES, 

General  Rule. — 1.  Write  the  quantities  to  be  added,  plac- 
ing those  that  are  similar  under  each  other. 

2.  Add  similar  terms,  and  annex  the  others,  with  their 
proper  signs. 

Review. — 55.  What  is  the  general  rule  for  the  addition  of  alge- 
braic quantities?  In  writing  them,  why  are  similar  quantities 
placed  under  each  other? 


22  RAY'S  ALGEBRA,  FIRST  BOOK. 

Remark. — It  is  not  absolutely  necessary  to  place  similar  terms 
under  each  other ;  but  as  we  can  add  only  similar  terms,  it  is  a  matter 
of  convenience  to  do  so. 

EXAMPLES. 

2.  Add  6a— 4c+36,  and  —  2a— 3c—  56. 

Ans.  4a — 7c— 26. 

3.  2a6-f  c,  4ax— 2c+14,  12— 2ax,  and  6a6+3c— x. 

Ans.  8a&+2aje+2c+26— x. 

4.  14a-{-x,  136—3/,  —  lla+2y,  and  —2a— 126+z. 

Ans.  a+6-}-#+ty+z- 

Note. — Since  the  quantities  in  parenthesis  are  to  be  considered 
as  one  quantity,  it  is  evident  that  3  times,  5  times,  and  7  times  any 
quantity  whatever=15  times  that  quantity. 

Add  together 

5.  2c(a»— 62),—  3c(a2— 62),  6c(a2— 62),  and— 4c(a2— 62). 

Ans.  c(a2 — 62). 

6.  Saz— Uy—  8,  —  2aH-56jH-6,  5a*+66#— 7,  and 
— -8cr* — t%-f-5.  Ans.  —  2az— 4. 

7.  8a  +  6,  2a— 6+c,  — 3a+56+2o7,  —  66— 3c+3d,  and 
-5a+7c— 2a1.  Ans.  2a— 6+5c-f-3a\ 

8.  *Jx — 6y-\-&z-\-& — (j,  — x — Sy — 8 — y,  — x-\-y — Sz — 1 
+7^,  —2x+3t/-t-3z—l—g,  and  a-ffy— 5*+9+£. 

Ans.  4aj+3y+2  +  5y. 

9.  5a362  —  8a2l*±x2y-{-xy2,  4a2bz—1a?b2  —  Zxy2+6x2y, 
3a362-j-3a263 — Sx2y-\-bxy2}  and  2a263 — a362 — 3a?y — Sxy2. 

Ans.  a2b*-\-x2y. 


SUBTRACTION 


5G.  Subtraction,  in  Algebra,  is  the  process  of  finding 
the  difference  between  two  algebraic  quantities. 

The  quantity  to  be  subtracted  is  called  the  subtrahend; 
that  from  which,  the  subtraction  is  to  be  made,  the  minuend; 
the  quantity  left,  the  difference  or  remainder. 


SUBTRACTION.  23 

Remark. — The  word  subtrahend  means,  to  be  subtracted;  the 
word  minuend,  to  be  diminished. 

1.  Thomas  has  5a  cents;  if  he  give  2a  cents  to  his 
brother,  how  many  will  he  have  left? 

Since  5  times  any  quantity,  diminished  by  2  times  the  same  quan- 
tity, leaves  3  times  the  quantity,  the  answer  is  evidently  3a;  that 
is,  ba — 2a=Sa.     Hence, 

To  find  the  difference  between  two  positive  similar 
quantities, 

Find  the  difference  between  their  coefficients,  and  to  it  annex 
the  common  letter  or  letters. 

(2)  (3)  (4)  (5) 

From    hx  *7ab .  Sxy  lla2x 

Take    3a;  Sab  bxy  ba2x 

Remainder     2x  £ab  Sxy  6a'2x 

6.  From  9a,  take  4a ...     Ans.  5a. 

7.  From  116,  take  116 Ans.  0. 

8.  From  3a2,  take  2a2 Ans.  a2. 

9.  From  7b2xy,  take  4b2xy Ans.  Sb2xy. 

57. — 1.  Thomas  has  a  apples;  if  he  give  away  b  apples, 
what  expression  will  represent  the  number  he  has  left  ? 

If  a  represents  6,  and  b  4,  the  number  left  will  be  represented 
by  6 — 4,  or  2;  and  whatever  numbers  a  and  b  represent,  it  is  evi- 
dent that  their  difference  may  be  expressed  in  the  same  way ;  that 
is,  by  a — b.     Hence, 

To  find  the  difference  between  two  quantities  not  similar, 
Place  the  sign  minus  before  the  quantity  to  be  subtracted. 

Observe  that  the  sign  of  the  quantity  to  be  subtracted  is 
changed  from  plus  to  minus. 

Review. — 56.  What  is  Subtraction,  in  Algebra?  What  is  the  quan- 
tity to  be  subtracted  called?  The  quantity  from  which  the  subtrac- 
tion is  to  be  made?     What  does  subtrahend  mean?     Minuend? 

56.  How  find  the  difference  between  two  positive  similar  quan- 
tities?    57.  How  between  two  quantities  not  similar? 


24  RAY'S  ALGEBRA,  FIRST  BOOK. 

2.  From  c,  take  d Ans.  c — d. 

3.  From  2m,  take  3n Ans.  2m — 8n. 

4.  From  a2x,  take  ax2 Ans.  a2x — ax2. 

5.  From  x2,  take  cc Ans.  x2 — x. 

58. — 1.  Let  it  be  required  to  subtract  5-J-3  from  9. 

If  we  subtract  5  from  9,  the  remainder  will  be  9 — 5 ;  but  we  wish 
to  subtract,  not  only  5,  but  also  3.     Hence, 

After  we  have  subtracted  5,  we  must  also  subtract  3 ;  this  gives 
for  the  remainder,  9 — 5 — 3,  which  is  equal  to  1. 

2.  Let  it  be  required  to  subtract  5 — 3  from  9. 

If  we  subtract  5  from  9,  the  remainder  is  9—5;  but  the  quantity 
to  be  subtracted  is  3  less  than  5.     Hence, 

We  have  subtracted  3  too  much;  we  must,  therefore,  add  3  to  9 — 5, 
which  gives  for  the  true  remainder,  9 — 5-J-3,  or  7. 

3.  Let  it  now  be  required  to  subtract  b — c  from  a. 

If  we  take  b  from  a,  the  remainder  is  a — 6;  but,  in  doing  this,  we 
have  subtracted  C  too  much ;  hence,  to  obtain  the  true  result,  wo 
must  add  C.     This  gives  the  true  remainder,  a — b-\-c. 

If  a=9,  6=5,  and  c=3,  the  operation  and  illustration  by  figures 
would  stand  thus:     From  a  from  9  =9 

Take   b—G  take  5—3  =2 

Remainder,  a — 6-|-c         Rem.     9 — 5-|-3  =7 

For  further  illustration,  take  the  following  : 

4.  a — (c — a)  =a — c-j-a  =2a — c. 
a — (a — c)  =a — a-\-c  =c. 
a+h—ta—b)  =a-±b—a+b  =2b. 

Observe  that  in  each  of  the  preceding  examples,  the 
signs  of  the  subtrahend  are  changed  from  plus  to  minus 
and  from  minus  to  plus.     Hence 


TO    FIND    THE    DIFFERENCE    BETWEEN     TWO     ALGEBRAIC 
QUANTITIES, 
Rule. — 1.  Write  the  quantity  to  be  subtracted  under  that 
from  which  it  is  to  be  taken,  placing  similar  terms  under 
each  other. 


SUBTRACTION. 


25 


2.  Conceive  the  signs  of  all  the  terms  of  the  subtrahend  to 
be  changed,  and  then  reduce  by  the  ride  for  Addition. 

Note. — Beginners  can  write  the  example  a  second  time,  then 
actually  change  the  signs,  find  add,  as  in  the  following  example, 
until  they  become  familiar  with  the  rule. 


From  5a-\-3b — c 

The  same  with  the 

5a-j-36— c 

Take  2a— 2b— 3c 

signs  of   the  subtra- 
hend changed. 

— 2a-|-26+3c 

Item.       3a-\-5b-\-2c 

3a_j_56_|_2c 

EXAMPLES. 

(6) 

(<) 

(8) 

From  Sax — 2y 

4tcx2 — 3by2          Sxyz-\-3az — 8 

Take  2ax-\-3y 

2cx — 3by2          bxyz — 3a^-j-8 

Rem.       ax — by 

4  ex2 — 2  ex           Sxyz  -\-6az — 16 

(«) 

(10) 

(11) 

From  7x+4y 

3a— 26              6ax— 4^+3 

Take    6x — y 

5a— 3b              3ax— 6^+2 

12.  From  14,  take  ah— 5 

Ans.  19 — ab. 

13.  From  a-j-6,  take  a 

.     .       Ads.  b. 

14.   From  a,  take 

a-ffc 

.     .  Ans.  — b. 

15.  From  x,  take  r 

r— 5 

.     .      Ans.  5. 

16.  From  x-\-y,  take  x — y.      .     .     . 

.     .     Ans.  2y. 

17.  From  x — y,  take  x-\-y.      .     .     . 

.      Ans.  — 2y. 

18.  From  x — y,  take  y — x.       .     .     . 

Ans.  2x — 2y. 

19.  From  x-\-y-\-z 

take  x — y — z. 

Ans.  2y+2z. 

20.  From  bx+3y- 

—z,  take  4x-\-3y-\-z. 

.   Ans.  x — 2z. 

21.  From  a,  take  • 

— a, 

.     .     Ans.  2a. 

22.  From  8a,  take 

—3a 

.     .  Ans.  11a. 

23.  From  56,  take  116 

.      Ans.  — 6b. 

24.  From  a,  take 

—b 

Ans.  a+6. 

25.  From  —9a,  ta 

ke  3a 

.  Ans.  —12a. 

26.  From  —7a,  take  —7a 

.     .      Ans.  0. 

27.  From  —6a,  ta 

ke  — 5a 

.     .  Ans.  — a. 

1st  Bk.              3 

26  RAYS  ALGEBRA,  FIRST  BOOK. 

28.  From  —13,  take  3 Ans.  —16. 

29.  From  —9,  take  —16: Ans.  7. 

30.  From  12,  take  —8 Ans.  20. 

31.  From  —14,  take  —5 Ans.  —9. 

32.  From  13a—2b  +  9c—3d,  take  8a—6b+9c—10d 
+  12.  Ans.  5a-{-4b  +  7d-12. 

33.  From  —  *Ja-\-3m—  8x,  take  —  6a—  bin— 2x-\-Zd. 

Ans.  — a-\-8m — 6x — 3d. 

34.  From  6a  +  5— 36,  take  —2a— 9b— 8. 

Ans.   8a+6Z>+13. 

35.  From  3a# — 2#2,  take  — 5acc — 8y2.     Ans.  8aa;+6y. 

36.  From  \xhf—  5cz-f-8m,  take  — cz-\-2xhf — 4cz. 

Ans.  2#2iy3-f-8»i. 

37.  From  ar3 — llaryz-f-3a,  take  — 6#y2-f-7 — 2a — bxyz. 

Ans.  #3-j-5a — 7. 

38.  From  50-r-y),  take  2(a?+y).      .     .     Ans.   3  (>-}-#). 

39.  From  3a(x — z),  take  a(x — z)      .     .    Ans.  2a(x—  z). 

40.  From  7a2(c— z)—  ab(c— d),  take  ba\c—z)—bab 
(c—d).  Ans.  2a2(c— z)+ ±ab{c— d). 

«50.  It  is  sometimes  convenient  to  simply  indicate  the 
subtraction  of  a  polynomial.  This  may  be  done  by  in- 
closing it  in  a  parenthesis,  and  then  placing  the  sign  minus 
before  it. 

Thus,  to  subtract  a — b  from  2a,  write  it  2a — (a — 6),  which  is 
equal  to  2a — a-f-6,  and  reduces  to  a-\-b. 

By  this  transformation,  the  same  polynomial  may  be  written  in 
several  different  forms,  thus: 

a— 6-f-c— d=a— b— (d— c)=a— d— ( b—c)=a—(  b—c-\-d) . 

In  the  following  examples,  introduce  all  the  quantities 
except  the  first  into  a  parenthesis,  and  precede  it  by  the 
sign  minus,  without  altering  the  value  of  the  expressions. 

Review. — 58.  In  subtracting  b — c  from  a,  after  taking  away  b, 
have  we  subtracted  too  much,  or  too  little?  What  must  be  added,  to 
obtain  the  true  result?  Why?  What  is  the  general  rule  for  find- 
ing the  difference  between  two  algebraic  quantities? 

59.  How  can  the  subtraction  of  an  algebraic  quantity  be  indicated? 


SUBTRACTION.  27 

1.  a — b-\-c Ans.  a — (o — c). 

2.  l-\-<— d Ans.  b— (d—  c). 

3.  ax-\-bc — cd-\-h Ans.  ax — (cd — be — K). 

4.  m — n — z — s Ans.  m — (ii-\-z-\-s). 

Let  the  pupil  take  the  preceding  polynomials,  and  write 
them  in  all  possible  modes,  by  including  either  two  or  more 
terms  in  a  parenthesis. 


OBSERVATIONS    ON   ADDITION   AND   SUBTRACTION. 

©O.  It  has  been  shown  that  algebraic  addition  is  the  process 
of  collecting  two  or  more  quantities  into  one. 

If  these  quantities  are  either  all  positive  or  all  negative,  the  sum 
will  be  greater  than  either  of  the  individual  quantities. 

If  some  of  the  quantities  are  positive  and  others  negative,  the 
aggregate  may  be  less  than  either  of  them,  or  it  may  be  nothing. 

Thus,  the  sum  of  -f4a  and  — 3a,  is  a;  while  that  of  -\-a  and  — a, 
is  zero,  or  0. 

As  the  introduction  of  the  minus  sign  makes  the  operations  of 
algebraic  addition  and  subtraction  differ  materially  from  those  of 
arithmetic,  it  will  be  proper  to  enter  into  a  further  explanation 
of  them. 

Ol.  In  arithmetical  addition,  when  we  say  the  sum  of  5  and 
3  is  8,  we  mean  that  their  sum  is  8  greater  than  0. 

In  Algebra,  when  we  say  the  sum  of  5  and  — 3  is  2,  we  mean 
that  the  aggregate  effect  of  adding  5  and  subtracting  3,  is  the  same 
as  that  of  adding  2.  "When  we  say  the  sum  of  — 5  and  -\-S  is  — 2, 
we  mean  that  the  result  of  subtracting  5  and  adding  3,  is  the  same 
as  that  of  subtracting  2. 

Some  say  that  numbers,  with  a  negative  sign,  such  as  — 3,  repre- 
sent quantities  less  than  nothing.  This  phrase,  however,  is  objection- 
able. If  we  understand  by  it  that  any  negative  quantity,  added  to 
a  positive,  will  produce  a  result  less  than  if  nothing  had  been  added 
to  it;  or,  subtracted,  will  produce  a  result  greater  than  if  nothing  had 
been  taken  from  ft,  then  the  phrase  has  a  correct  meaning.     Thus, 

Review. — 60.  When  is  the  sum  of  two  algebraic  quantities  less 
than  ekher  of  them?     When  equal  to  zero? 


28  RAY'S  ALGEBRA,  FIRST  BOOK. 

If  we  take  any  number,  as  10,  and  add  to  it  the  numbers 
3,  2,  1,  0,  — 1,  — 2,  and  — 3,  it  will  be  seen  that  adding  a  negative 
number  produces  a  less  result  than  adding  zero. 

10  10  10  10  10  10  10 

3  2  1  0—1—2-3 

13  12  11  10  9  8  7 

Hence,  adding  a  negative  number  produces  the  same  result  as 
subtracting  an  equal  positive  number. 

Again,  if  from  any  number,  as  10,  we  subtract  3,  2,  1,  0,  — 1, 
— 2,  and  — 3,  it  will  be  seen  that  subtracting  a  negative  number 
produces  a  greater  result  than  subtracting  zero: 

10  10  10  10  10  10  10 

3  2  1  0—1—2—3 

~7  ~8  ~9  10  11  12  13 

Hence,  subtracting  a  negative  number  produces  the  same  result 
as  adding  an  equal  positive  number. 

62.  In  consequence  of  the  results  they  produce,  it  is  custom- 
ary to  say,  of  negative  algebraic  quantities,  that  those  which  are 
numerically  the  greatest  are  really  the  least.  Thus,  — 3  is  less  than 
— 2,  though  numerically  greater. 

©3.  A  correct  idea  of  this  subject  may  be  gained  by  consider- 
ing such  questions  as  the  following: 

How  will  the  money  in  a  drawer  be  affected,  if  $20  are  taken  out, 
afterward  $15  put  in,  after  this  $8  taken  out,  and  then  $10 put  in? 

Or,  in  other  words,  what  is  the  sum  of  — 20,  -[—15,  — 8,  and  -j-10? 

The  answer,  evidently,  is  — 3;  that  is,  the  result  of  the  whole 
operation  diminishes  the  money  in  the  drawer  $3. 

Had  the  answer  been  positive,  the  result  of  the  operation  would 
have  been  an  increase  of  the  amount  of  money  in  the  drawer. 

Again,  suppose  latitude  north  of  the  equator  to  be  reckoned  -}-, 
and  that  south  — ,  in  the  following  question : 

A   ship,    in    latitude   10   degrees    north,    sails   5   degrees   south, 

Review. — 01.  What  is  meant  by  saying  that  the  sum  of  -{-5  and 
—3,  is  -{-2?     That  the  sum  of  —5  and  -f- 3,  is  —2? 

01.  Is  it  correct  to  say  that  any  quantity  is  less  than  nothing? 
What  is  the  effect  of  adding  a  positive  quantity?  A  negative  quan- 
tity?    Of  subtracting  a  positive  quantity?     A  negative  quantity? 

02.  In  comparing  two  negative  algebraic  quantities,  which  is 
least?     Which  numerically  greatest? 


SUBTRACTION.  29 

then  7  degrees  north,  then  9  degrees  south,  then  3  degrees  north ; 
what  is  her  present  latitude? 

Here,  -{-10,  —5,  -j-7,  —9,  and  -f-3,  are  evidently  -j-6;  that  is,  the 
ship  is  in  6  degrees  north  latitude. 

Had  the  sum  of  the  negative  numbers  been  the  greater,  the  ship 
would  have  been  in  south  latitude. 

Other  questions  of  a  similar  nature  will  readily  suggest  them- 
selves. 

G4.  Subtraction,  in  arithmetic,  shows  the  method  of  finding  the 
excess  of  one  quantity  over  another  of  the  same  kind. 

In  this  case,  the  subtrahend  must  be  lees  than  the  minuend,  and 
the  signs  are  regarded  as  the  same. 

In  algebraic  subtraction,  the  two  quantities  may  have  either  like 
or  unlike  signs,  and  the  difference  is  often  greater  than  either  of 
the  quantities.  To  understand  this  properly  requires  a  knowledge 
of  the  nature  of  positive  and  negative  quantities. 

All  quantities  are  to  be  regarded  as  positive,  unless,  for  some 
special  reason,  they  are  otherwise  designated.  Negative  quantities 
are,  in  their  nature,  the  opposite  of  positive  quantities. 

Thus,  if  a  merchant's  gains  are  positive,  his  losses  are  negative; 
if  latitude  north  of  the  equator  is  -|-,  that  south  is  — ;  if  distance 
to  the  right  of  a  certain  line  is  -}-,  that  to  the  left  is  — ;  if  elevation 
above  a  certain  point  is  -{-,  that  below  is  — ;  if  time  after  a  cer- 
tain hour  is  -j-,  time  before  that  hour  is  — ;  if  motion  in  one  direc- 
tion is  -{-,  motion  in  an  opposite  direction  is  — ;  and  so  on. 

With  these  illustrations  of  the  use  of  the  minus  sign,  it  is  easy 
to  see  how  the  difference  of  two  quantities,  having  the  same  sign, 
is  equal  to  their  difference;  and  also  how  the  difference  of  two  quan- 
tities, having  different  signs,  is  equal  to  their  sum. 

1.  One  place  is  situated  10,  and  another  6  degrees  north 
of  the  equator ;  what  is  their  difference  of  latitude  ? 

Here,  the  difference  between  -{-10  and  -{-6,  is  -f-4;  that  is,  the 
first  place  is  4  degrees  farther  north  than  the  second. 

2.  Two  places  are  situated,  one  10,  and  the  other  6  de- 
grees south  latitude  ;  what  is  the  difference  of  latitude  ? 

Review. — 64.  How  does  algebraic  differ  from  arithmetical  sub- 
traction? How  do  negative  quantities  differ  from  positive?  Illus- 
trate the  difference  by  examples. 


30  RAY'S  ALGEBRA,  FIRST  BOOK. 

Here,  the  difference  between  — 6  and  — 10,  is  — 4;  that  is,  the 
first  place  is  4  degrees  farther  south  than  the  second. 

3.  One  place  is  situated  10  degrees  north,  and  another  6 
degrees  south  latitude;  what  is  their  difference  of  latitude? 

Here,  we  are  to  find  the  difference  between  -]— 10  and  — 6,  or  to 
take  — 6  from  -{-10,  which,  by  the  rule  for  subtraction,  leaves  — |— 16 ; 
that  is,  the  first  place  is  16  degrees  north  of  the  other. 

Thus,  when  properly  understood,  the  results  of  algebraic  subtrac- 
tion are  always  capable  of  a  satisfactory  explanation. 


MULTIPLICATION. 

65.  Multiplication,  in  Algebra,  is  the  process  of  tak- 
ing one  algebraic  expression  as  many  times  as  there  are 
units  in  another. 

The  quantity  to  be  multiplied  is  called  the  multiplicand; 
the  quantity  by  which  we  multiply,  the  multiplier;  and  the 
result,  the  product. 

The  multiplicand  and  multiplier  are  called  factors. 

66.  Since  a,  taken  once,  is  represented  by  a,  taken  twice, 
by  a-\-a,  or  2a,  taken  three  times,  by  a-\-a-\-a,  or  3a.    Hence, 

To  multiply  a  literal  quantity  by  a  number, 

Rule. —  Write  the  multiplier  as  the  coefficient  of  the  literal 
quantity. 

1.  If  1  lemon  cost  a  cents,  what  will  5  lemons  cost? 

If  one  lemon  cost  a  cents,  five  lemons  will  cost  five  times  as  much; 
that  is-,  5a  cents. 

2.  If  1  orange  cost  c  cents,  what  will  6  oranges  cost? 

3.  Bought  a  pieces  of  cloth,  each  containing  b  yards, 
at  c  dollars  per  yard  ;  what  did  the  whole  cost? 

Review. — 65.  What  is  Multiplication,  in  Algebra?  What  the  mul- 
tiplicand? Multiplier?  Product?  What  are  the  multiplicand  and 
multiplier  generally  called?  66.  How  multiply  a  literal  quantity 
by  a  number?  . 


MULTIPLICATION.  31 

In  a  pieces,  the  number  of  yards  is  represented  by  ab,  or  ba,  and 
the  cost  of  ab  yards  at  C  dollars  per  yard,  is  represented  by  C  taken 
ab  times;  that  is,  by  abXc,  or  abc. 

©7.  It  is  shown  in  "Ray's  Arithmetic,  Third  Book,"  Art.  30, 
that  the  product  of  two  factors  is  the  same,  whichever  be  made  the 
multiplier.     Let  us  demonstrate  this  principle. 

Suppose  we  have  a  sash  containing  a  vertical,  and  b  horizontal 
rows;  there  will  be  a  panes  in  each  horizontal  row,  and  b  panes  in 
each  vertical  row ;  how  many  panes  in  the  window  ? 

The  number  of  panes  in  the  window  is  equal  to  the  number  in 
one  row,  taken  as  many  times  as  there  are  rows.  As  there  are  a  ver- 
tical rows,  and  b  panes  in  each  row,  the  number  is  represented 
by  b  taken  a  times;  that  is,  by  ab. 

Again,  since  there  are  b  horizontal  rows,  and  a  panes  in  each 
row,  the  whole  number  of  panes  is  represented  by  a  taken  b  times; 
that  is,  by  ba. 

As  ab  and  ba  each  represents  the  same  number,  it  follows  that 
ab=ba.     Hence, 

TJie  product  of  two  factors  is  the  same,  whichever  be  made 
the  mirttiplier. 

By  taking  a=3  and  6=4,  the  figure  in  the  margin 
may  be  used  to  illustrate  this  principle. 

So,  the  product  of  three  or  more  quantities  is  the 
same,  in  whatever  order  taken. 

Thus,  2X3X4=3X2X4=4X2X3,  since  the  product 
in  each  case  is  24. 

1.  What  will  2  boxes,  each  containing  a  lemons,  cost 
at  b  cents  per  lemon  ? 

One  box  will  cost  ab  cents,  and  2  boxes  will  cost  twice  as  much 
as  1  box;   that  is,  lab  cents. 

2.  What  is  the  product  of  lb,  multiplied  by  3a  ? 

The  product  will  be  represented  by  2b\Sct,  or  by  3«X26,  or  by 
2x3X#A  since  the  product  is  the  same,  in  whatever  order  the  fac- 
tors are  placed.     But  2x3=6;  hence,  2X^Xao=Qao- 

Review. — 67.  Prove  that  8  times  4  is  the  same  as  4  times  3.  That 
a  times  b  is  the  same  as  b  times  a.  Is  the  product  of  any  number  of 
factors  changed  by  altering  their  arrangement?  In  multiplying  one 
monomial  by  another,  how  is  the  coefficient  of  the  product  obtained? 


32  RAY'S  ALGEBRA,  FIRST  BOOK. 

Hence,  in  the  multiplication  of  one  monomial  by  another, 

The  coefficient  of  the  product  is  obtained  by  multiplying 
together  the  coefficients  of  the  multiplicand  and  multiplier. 
This  is  termed  the  Rule  of  the  Coefficients. 

©&.  If  we  take  any  two  factors,  as  2X3,  and  multiply  either  by 
any  number,  as  5,  the  products  will  be  10x3,  or  2x15,  either  of 
which  is  equal  to  30,  which  is  the  true  answer.     Hence, 

When  either  of  the  factors  of  a  product  is  multiplied,  the 
product  itself  is  multiplied. 

69. — 1.  What  is  the  product  of  a  by  a? 

As  bya=.ab,  so  aX«  would  be  written  aa;  but  this,  Art  33,  for 
brevity,  is  written  a2. 

2.  What  is  the  product  of  a2  by  a  ? 

Since  a2=aa,  the  product  of  a2  by  a  may  be  expressed  thus, 
aay^a,  or  aaa,  which  is  written  a?>.     Hence, 

The  exponent  of  a  letter  in  the  product  is  equal  to  the  sum 
of  its  exponents  in  the  two  factors. 

This  is  termed  the  Rule  of  the  Exponents. 

3.  What  is  the  product  of  a2  by  a2  ?     Ans.  aaaa,  or  a4. 

4.  Of  a2b  by  ab? Ans.  aaabb,  or  a*b2. 

5.  Of  2ab2  by  Sab?  ....     Ans.  6aabbb,  or  6d-V. 

TO   MULTIPLY  ONE    POSITIVE    MONOMIAL    BY  ANOTHER, 

Rule. — 1.  Multiply  the  coefficients  of  the  two  terms  together. 

2.  To  this  product,  annex  all  the  letters  in  both  quantities. 

3.  Wlien  the  same  letter  occurs  in  both  factors,  add  its  ex- 
ponents for  the  exponent  of  the  product. 

Note. — Write  the  letters  in  the  order  of  the  alphabet;  thus, 
aby^c^zabc. 

6.  Multiply  ab  by  x .    Ans.  abx. 

Review. — 68.  If  you  multiply  one  of  the  factors  of  a  product,  how 
does  it  affect  the  product?  69.  How  may  the  product  of  a  by  a  be 
written?     Of  a?  by  a? 


MULTIPLICATION. 


33 


7.  Multiply  2bc  by  mn.    . 

8.  Multiply  4ab  by  bxy.  . 

9.  Multiply  6by  by  3«#.  . 

10.  Multiply  Sa2b  by  4a6. 

11.  Multiply  2xy2  by  3afy.  • 

12.  Multiply  4afe*a:  by  bax2y. 

13.  Multiply  7^22  by  8afyz. 


Ans.  2hcmn. 
.  Ans.  20afory. 
.  Ans.  18afory. 
.  Ans.  12a»6*. 
Ans.  6ai3#3. 
Ans.  20aWy. 
.Ans.  56ar*y«2. 


Note. — Distinguish  carefully  between  the  coefficient  and  the  ex- 
ponent.    To  fix  this  in  the  mind,  answer  the  following  questions : 

What  is  2a — a2  equal  to,  when  a  is  1  ?  .  .  Ans.  1. 
"What  is  a2 — 2a  equal  to,  when  a  is  5  ?  .  .  Ans.  15. 
What  is  a3 — 3a  equal  to,  when  a  is  4?    .     .     Ans.  52. 

TO. — 1.  If  5  oranges  were  purchased  at  4  cents  apiece, 
and  2  lemons  at  the  same  price  ;  what  did  the  whole  cost  ? 

The  5  oranges  cost  20  cents,  the  2  lemons  cost  8  cents,  and  the 
whole  cost  was  20-J-8— 28  cents. 

The  work  may  be  written  thus :     5-J-2 

4 

20+8=28  cents. 

If  you  purchase  a  oranges  at  c  cents  apiece,  and  b  lemons  at  c 
cents  apiece,  what  is  the  cost  of  the  whole? 

The  cost  of  a  oranges  at  c  cents  each,  is  ae  cents;  the  cost 
of  6  lemons  at  c  cents  each,  is  be  cents,  and  the  whole  cost  is 
ac-{-bc  cents. 

The  work  may  be  written  thus;     a-f-6 

c 


ac-\-bc 
Hence,  when  the  sign  of  each  term  is  positive, 


TO    MULTIPLY    A    POLYNOMIAL    BY    A    MONOMIAL, 

Rule. — Multiply  each  term  of   the   multiplicand  by  the 
multiplier. 

Note. — It  is  most  convenient  to  place  the  multiplier  on  the  left 


34 


RAYS  ALGEBRA,  FIRST  BOOK. 


.     .  Ans.  ah-\-bd. 

Ans.  12e/#+15ay. 

.  Ans.  3wm-|-6n2. 

Ans.  a?y-\-xyz. 

Ans.  2abx2-\-habxy. 

Ans.  §7?z-\-kx2z2. 


EXAMPLES. 

2.  Multiply  a-\-d  by  b. .     . 

3.  Multiply  4z-f-5y  by  3a. 

4.  Multiply  m+2n  by  3ra. 

5.  Multiply  x2-\-y2  by  #y.    . 

6.  Multiply  2x+5y  by  a&r. 
T.  Multiply  3x2+2xz  by  2aw. 
8.  Multiply  ab-\-ax-\-xy  by  afocy. 

Ans.  a2b2xy-]-a2bx2y-]-abx2y2. 

•71o — 1.  Required  the  product  of  x-\-y  by  a-\-b. 
Here,  the  multiplicand  is  to  be  taken  as  many  times  as  there  are 
units  in  a+6,  and  the  whole  product  will  equal  the  sum  of  the  two 
partial  products.     Thus, 
x+y 
a+b 

ax-\-a$/—the  multiplicand  taken  a  times. 

bx-\-by=the  multiplicand  taken  b  times. 

CtX-\-ay+bx+by=the  multiplicand  taken  (a+6)  times. 
If  #=5,  2/— 6,  a=2}  and  6=3,  the  multiplication  may  be  arranged 
thus:     5+6 

2+3 

10+12=the  multiplicand  taken  2  times. 

15+18=rthe  multiplicand  taken  3  times. 

10+27+18=55=the  multiplicand  taken  5  times. 
Hence,  when  all  the  terms  in  each  are  positive, 


TO    MULTIPLY   ONE   POLYNOMIAL   BY   ANOTHER, 

Rule. — Multiply  each  term  of  the  multiplicand  by  each 
term  of  the  multiplier,  and  add  the  products  together. 

(2)  (3) 

a-\-b  .  a2b-\-cd 

a-\-b  ab-\-cd2 


!+a6 


ab+b2 


a2+2ab-{-b2 


a3b2-\-abcd 
aW-j-d^bcd'-t-abcd+chl3 


MULTIPLICATION.  85 

4.  Multiply  a-\-b  by  c-\-d.  .     .    Ans.  ac-\-ad-\-bc-\-bd. 

5.  2x+%  by  3«+26.     Ans.  6ax+9ay+4:bx+6bi/. 

6.  2a+36  by  3c-M.        Ans.  6ac+96c+2ad+36cZ. 

7.  m+7i  by  x-\-z Ans.  mx-\-iix-\-mz-]-nz. 

8.  4a  +  36by  2a  +  b.      .     .        Ans.  8«2+10^+362. 

9.  4a;+  5y  by  2a+8x. 

Ans.  8aa+10ay+12a:2+15a!y. 

10.  3^+2^  by  2x-\-3y.     .     .     Ans.  6x2+ 13^+6/. 

11.  a2+62  by  a+b Ans.  a3+trb+ab'2+b3. 

12.  3a2+262  by  2a2-{-362.      .    Ans.  6a*+l3aW+6b\ 

13.  a2+ab+b2  by  o-ffi.      .     Ans.  a*+2ci2b+2ab2-\-b\ 

14.  c3+J3  by  c-f-rf Ans.  c4-f-ca3+c3</-|-d4. 

15.  x2-\-2xj-\-yl  by  ^-j-y.  .      Ans.  a^+  3^+3 xy'^if. 


OF    THE    SIGNS. 

T2-  In  the  preceding  examples,  it  was  assumed  that 
the  product  of  two  positive  quantities  is  positive.  This, 
and  the  other  possible  cases,  may  be  proved,  as  follows : 

1st.  Let  it  be  required  to  find  the  product  of  -f  b  by  a. 

The  quantity  b  taken  once,  is  -|-6;  taken  twice,  is  +26;  taken 
3  times,  is  +36,  and  so  on.     Therefore,  taken  a  times,  it  is  -\-ab. 

Hence,  the  product  of  two  positive  quantities  is  positive;  or,  more 
briefly,  plus  multiplied  by  plus  gives  ^Zws. 

2d.  Let  it  be  required  to  find  the  product  of  — b  by  a. 

The  quantity  — 6  taken  once,  is  — 6;  taken  twice,  is  — 26;  taken 
3  times,  is  — 36;  taken  a  times,  is  — ab.     Hence, 

A  negative  quantity  multiplied  by  a  positive  quantity,  gives  a  nega- 
tive product;  or,  minus  multiplied  by  plus  gives  minus. 

3d.  Let  it  be  required  to  multiply  b  by  — a. 

Review. — To  what  is  the  exponent,  of  a  letter  in  the  product 
equal?     Rule  for  multiplying  one  positive  monomial  by  another. 

70.  What  is  the  product  of  tf+6,  by  cl  When  the  signs  are  posi- 
tive, how  multiply  a  polynomial  by  a  monomial?  71.  How  two 
polynomials  ? 


36  RAY  S  ALGEBRA,  FIRST  BOOK. 

When  the  multiplier  is  positive,  we  understand  that  the  multipli- 
cand is  to  be  added  to  0  as  many  times  as  there  are  units  in  the  mul- 
tiplier. Now,  since  the  negative  sign  always  expresses  the  opposite 
of  the  positive  sign,  when  the  multiplier  is  negative  the 'multipli- 
cand must  be  subtracted  from  0  as  many  times  as  there  are  units  in 
the  multiplier. 

The  quantity  b  subtracted  once,  is  — b ;  subtracted  twice,  it 
is  — 26;  subtracted  a  times,  it  is  —  ab.     Hence, 

A  positive  quantity  multiplied  by  a  negative  quantity,  gives  a 
negative  product;  or,  plus  multiplied  by  minus  gives  minus.  ■ 

4th.  Let  it  be  required  to  multiply  — b  by  — a. 

According  to  the  principle  stated  above,  — b  is  to  be  subtracted 
from  0  a  times;  subtracted  once,  it  is  +6;  subtracted  twice,  it 
is  +26;  subtracted  a  times,  it  is  -\-ab.     Hence, 

The  product  of  two  negative  quantities  is  positive;  or,  minus  multi- 
plied by  minus  gives  plus. 

Note. — The  following  proof  of  the  last  principle  is  generally 
regarded  as  more  satisfactory  than  the  preceding: 

To  find  the  product  of  c—d  by  a—b. 

Here,  it  is  required  to  take  c — d  a  times,  and  then  subtract 
from  this  product,  C — d  taken  b  times. 

The  multiplication  of  c—d  by  a — b  may  be  written  thus : 

c—d 

a—b 


ac—ad=c — d  taken  a  times. 

-{-be — bd—c — d  taken  b  times.    Subtract  from  the  above. 

ac—ad — bc-\-bd,  the  true  product. 

Observing  the  answer,  which  we  know  to  be  the  true  product,  we  see 
that  +cX+«  must  give  -\-ac,  — dx+a=— °^>  +cy(—b=—bc, 
and  —dX—b=+bd;  which  last  result  is  the  thing  to  be  proved. 
To  illustrate  by  figures,  find  the  product  of  7 — 4  by  5 — 3. 

We  first  take  5  times  7—4;  this  gives  a  prod- 
— *  uct  too  great,  by  3  times  7—4,  or  21—12,  which 

being  subtracted  from  the  first  product,  gives 


35—20  for  the  true  result,  35—41+12,  which  reduces 

+21—12       to  +6.     This  is  evidently  correct,  for  7—4=3, 

35 41_|_12       and  5—3=2,  and  the  product  of  3  by  2  is  6. 

Hence, 


MULTIPLICATION. 


37 


THE     GENERAL     RULE, 

FOR    THE    SIGNS. 

1.  Plus  multiplied  by  p>lus,  or  minus  by  minus,  gives  plus. 

2.  Plus  multiplied  by  minus,  or  minus  by  plus,  gives  minus. 

3.  Or,  the  product  of  like  signs  gives  plus,  and  of  unlike 
signs  gives  minus. 

From  all  the  preceding,  we  derive  the  following 

GENERAL    RULE, 
FOR  THE   MULTIPLICATION   OF  ALGEBRAIC  QUANTITIES. 

1.  Beginning  at  the  left  hand,  multiply  each  term  of  the 
multiplicand  by  each  term  of  the  multiplier,  observing  that 
like  signs  give  plus  and  unlike  signs  give  minus. 

2.  Add  the  several  partial  products  together. 

NUMERICAL    EXAMPLES, 
TO    VERIFY    THE    RULE    OF    THE    SIGNS. 

1.  Multiply  8—3  by  5.  ... '",     Ans.  40—15=25. 

2.  Multiply  9—5'  by  8—2.      .  Ans.  72—58+10=24. 

3.  Multiply  8— 7  by  5—3.       .     Ans.  40— 59+ 21=2. 


GENERAL     EXAMPLES. 


1.  Multiply  Ba2xy  by  laxf  . 

2.  Multiply  —  ba2b  by  Sab*. . 

3.  Multiply  — bx2y  by  — bxy2 

4.  Multiply  3a— 2b  by  4c.    . 

5.  Multiply  3^+2^  by  —  2x. 

6.  Multiply  a-\-b  by  x — y.     . 


.  Ans.  21a3x'1y4. 
.Ans.  —  IbaW. 
Ans.  25ary. 
Ans.  12«c — Sbc. 
Ans.  — 6x2 — 4xy. 
Ans.  ex — ay-\-bx — by. 


Review.— 72.  What  is  the  product  of  -f  6  by  +a?  Why?  The 
product,  of  —b  by  al  Why?  The  product  of  +  6  by  —a?  Why? 
The  product  of  —3  by  —2  ? 

72.  What  does  a  negative  multiplier  signify?  What  does  minus 
multiplied  by  minus  produce  ?  General  rule  for  the  signs?  For 
the  multiplication  of  algebraic  quantities? 


38  RAY'S  ALGEBRA,  FIRST  BOOK. 

7.  Multiply  a— b  by  a—b Ans.  a2—2ab~\-b2. 

8.  Multiply  a2-\-ac-\-  C*  by  a — c Ans.  a3 — c3. 

9.  Multiply  m-\-n  by  m — n Ans.  m2 — n'2. 

10.  Multiply  a2—2ab-\-b2hya+b.  Ans.  a3— a-fl— aV^-b\ 

11.  Multiply  3afy— Ixf+if  by  2xy+y2. 

Ans.  6#y-f  3^-y3— 4^/+/. 

12.  Multiply  as-J-2c5-f^  by  a2—  2ab-\-b2. 

Ans.  a4—  2cr62-f  bK 

13.  Multiply  7/2—3/+ 1  by  y-fl Ans.  y3-f  1. 

14.  Multiply  x'2-\-y2  by  x2 — y2 Ans.  #4— y. 

15.  Multiply  a2— 3a+8  by  a-f- 3..     .   Ans.  a3— a-f-24. 

16.  Multiply  2x2 — 3xy-\-y2  by  x2 — hxy. 

Ans.  2a:4— 13afy+l 6x2y2—oxy\ 

17.  Multiply  3.H-55  by  3a— bb.   .     .     Ans.  9a2— 25&2. 

18.  Multiply  2a2-4az+2z2  by  3a— 3x. 

Ans.  6a3— 18a=*+l&a?2— 6s\ 

19.  Multiply  5^+%3  ]jy  5^— 3v/3.    .    Ans.  25*6— 9/. 

20.  Multiply  2«3+2a2*+2«:r2-h2;r5  by  3a—  3x. 

Ans.  6a4 — 6#4. 

21.  Multiply  3a2+ 3a*+3.e2  by  2«2-2^. 

Ans.  6a* — Qax*. 

22.  Multiply  3a2+5ax— 2x2  by  2a— x. 

Ans.  6a3-j-7«'-'z—  §ax2+ 2x\ 

23.  Multiply  x*-\-xx-\-x2  by  x2 — 1 Ans.  Xs — x2. 

24.  Multiply  ar-f  xy-{-y2  by  V2  — xy+y2.  Ans.  #4-f  #2/-f-?/4. 

25.  Multiply  a3-f-tt2o-f«o2-f 63  by  a—b.     .     Ans.  «4— b\ 

In  the  following,  multiply  together  the  quantities  in  the 
parentheses. 

26.  (x—3)(x—  3)(x— 3).  .     .     Ans.  x"—9x2-^2lx—2l. 

27.  (a— 4)(s— 5)(a?+4)(ar+5).       Ans.  *4— 41ar+400. 

28.  (a-fc5(»— c)(a-j-c)(a— e).  .     .     Ans.  a4— 2«2c2-f c4. 
99.  (a2+^+c2_a5_ac_^,^a_|_5_j_r^ 

Ans.  r/3-f  63-f-c3— 3r<Sr. 
30.  (/i2-|-,i+l)(^_|_n+i)(?l_i)(?i_iy 

Ans.  n6— 2/i3-f-l. 


DIVISION. 


DIVISION. 

73.  Division,  in  Algebra,  is  the  process  of  finding  how 
many  times  one  algebraic  quantity  is  contained  in  another. 

Or,  having  the  product  of  two  factors,  and  one  of  them 
given,  Division  teaches  the  method  of  finding  the  other. 

The  number  by  which  we  divide  is  called  the  divisor; 
the  number  to  be  divided,  the  dividend;  the  number  of 
times  the  divisor  is  contained  in  the  dividend,  the  quotient. 

74.  Since  the  divisor  is  the  known  factor  and  the  quo- 
tient the  one  found,  their  product  must  always  be  equal 
to  the  dividend. 

Division  may  be  indicated  by  writing  the  divisor  under 
the  dividend  in  the  form  of  a  fraction,  or  as  in  arithmetic. 

Thus,  ab  divided  by  a,  is  written  — ,  or  a)ab. 

a 

Note  — In  solving  the  following,  give  the  reason  for  the  answer, 
as  in  the  solution  to  the  first  question. 


1.  How  many  times  is  x  contained  in  4x?      Ans. 


4x 


4x  divided  by  x,  equals  4,  because  the  product  of  4  by  X  is  Ax. 


2.  How  many  times  is  a  contained  in  6a  ? 

3. 

4. 

5. 
6. 

1. 


9. 
10. 


Is  a  contained  in  ab?      .     . 
Is  b  contained  in  Sab  ?    .     . 
Is  2  contained  in  4a  ?      .     . 
Is  2a  contained  in  4ab  ? 
Is  a  contained  in  a3  ?       . 
Is  ab  contained  in  ha2b  ? 
Is  4ab2  contained  in  12a3&3c? 
Is  2a2  contained  in  6a5b  ? 


.Ans.  6. 
.  Ans.  b. 
Ans.  3a. 
Ans.  2a. 
Ans.  2b. 
Ans.  a*. 
Ans.  5a. 
Ans.  3a2bc. 


Solution, 


6a56     G 


2a2 


_a>>-2&=3a36. 


Ans. 


Review. — 73.  What  is  Algebraic  Division?  The  divisor?  The 
dividend?  The  quotient?  74.  To  what  is  the  product  of  the  quo- 
tient and  divisor  equal?     Why?     How  is  division  indicated? 


40  RAYS  ALGEBRA,  FIRST  BOOK. 

In  obtaining  the  quotient,  in  the  foregoing  example,  we  readily  see, 

1st.  That  2  must  be  multiplied  by  3  to  produce  6. 

2d  That  a2  must  be  multiplied  by  a3  to  produce  a5;  or,  we  must 
subtract  2  from  5  to  find  the  exponent  of  a  in  the  quotient. 

3d.  That  since  b  is  in  the  dividend,  but  not  in  the  divisor,  it  must 
be  in  the  quotient,  so  that  the  product  of  the  divisor  and  quotient 
may  equal  the  dividend. 

75.  It  remains  to  ascertain  the  rule  for  the  signs. 

Since    -|~aX+^=+ttfy   — «X+&= — Clb,   -\-a)>(—b=—ab,  and 

-a/  —b=-\-ab, 

4-ab  — ab  — ab  ,  4-ab 

Therefore,       -^-=4- a,       -,-=—a,       -5-=4-a,  and       -=-= — a. 

'  -f  b       '     '  -f  b  '  —  b       '  —  6 

Or,  like  signs  give  plus,  and  unlike  signs  give  minus.     Hence, 


TO    DIVIDE    ONE    MONOMIAL    BY    ANOTHER, 

Rule. — 1.  Divide  the  coefficient  of  the  dividend  by  that 
of  the  divisor ;  observing,  that  like  signs  give  plus,  and  unlike 
signs  minus. 

2.  For  any  letter  common  to  the  divisor  and  dividend,  if 
it  has  the  same  exponent  in  both,  suppress  it;  if  riot,  subtract 
its  exponent  in  the  former  from  its  exponent  in  the  latter,  for 
its  exponent  in  the  quotient. 

3.  Annex  the  letters  found  in  the  dividend,  but  not  in  the 
divisor. 

Note. — The  pupil  must  recollect  that  a  is  the  same  as  a1. 


EXAMPLES. 

11.  Divide  IMbc  by  3a2b.     .     . 

12.  Divide  2lx2y2  by  —Zxy.  .     . 

13.  Divide  — 18a3.x  by  — Qax.    . 

14.  Divide  —12ci.r?y5  by  —  4c4a^2. 

15.  Divide  6acx2y6v  by  %ax2yH.  . 


.  .  Ans.  5a<\ 
.  Ans.  — 9xy. 
.  .  Ans.  3«2. 
Ans.  Sx2y3. 
.     .  Ans.  2cy2. 


Review. — 75.  When  the  signs  of  the  dividend  and  divisor  are 
Alike,  what  will  be  the  sign  of  the  quotient?  Why?  When  unlike? 
Why?     Rule  for  dividing  one  monomial  by  another? 


DIVISION.  41 

16.  Divide  — 10c2afy5y  by  — 2cy4v.  .     .     .      Ans.  bcxby. 

17.  Divide  —  28«c2;cVy2  by  Waxhf.    .     Ans.  —  2c2xH> 

18.  Divide  30«c4eVy2  by  —  2aex\  .      .   Ans.  — locV/ 

Note. — The  following  may  be  omitted  until  the  book  is  reviewed 

19.  Divide  (x-\-y)2  by  (x-\-y) Ans.  (x-\-y) 

20.  Divide  (a+by  by  (a  +  b) Ans.  (a-f-fc)3 

21.  Divide  6(m-H03  by  2(w+w).  .     .    Ans.  3(m-J-*)s 

22.  Divide  6a2^>(^+?/)3  by  2ai(a:+J/)2.    Ans.  3a(aH-y). 

23.  Divide  (.t — #)3(m — n)2  by  (a? — y)2(m — n)2. 

Ans.  (x^-y). 

TO.  It  is  evident  that  one  monomial  can  not  be  divided 
by  another,  in  the  following  cases  : 

1st.  When  the  coefficient  of  the  dividend  is  not  exactly 
divisible  by  the  coefficient  of  the  divisor. 

2d.  When  the  same  literal  factor  has  a  greater  exponent 
in  the  divisor  than  in  the  dividend. 

3d.  When  the  divisor  contains  one  or  more  literal  fac- 
tors not  found  in  the  dividend. 

In  each  of  these  cases,  the  division  is  indicated  by  writ- 
ing the  divisor  under  the  dividend,  in  the  form  of  a  fraction. 
This  fraction  may  then  be  reduced,  Art.  129. 

TT.  It  has  been  shown,  Art.  68,  that  any  product  is 
multiplied  by  multiplying  either  of  its  factors ;  hence, 
conversely,  any  dividend  will  be  divided  by  dividing  either 
of  its  factors. 


Thus,  ^=2X6=12;  or, -^-=4X3=12 


T8.  In  multiplying  a  polynomial  by  a  monomial,  wc 
multiply  each  term  of  the  multiplicand  by  the  multiplier. 
Hence, 

Review. — 76.  In  what  cases  is  the  exact  division  of  one  mono- 
mial by  another  impossible?  78.  Rule  for  dividing  a  polynomial 
by  a  monomial?  ^ — 

UtBk.  4*   >*^>l«^^ 

f  OF  THE 

f    UNIVERSITY   ) 


42 


TO   DIVIDE   A   POLYNOMIAL   BY   A   MONOMIAL, 


Rule. — Divide  each  term  of  the  dividend  by  the  divisor, 
according  to  the  ride  for  the  division  of  monomials. 

Note  . — Place  the  divisor  on  the  left,  as  in  arithmetic. 


1.  Divide  6x+12y  by  3.      .     .     :     .     .  Ans.  2*+4y. 

2.  Divide  15^—206  by  5 Ans.  &»— 4'ft 

3.  Divide  21a-|-35Z>  by  — 7.     .     .     .  Ans.  —3a— 5&. 

4.  Divide  abc — acf  by  ac Ans.  b — f. 

5.  Divide  lOax — lhay  by  — ha.    .     .  Ans.  — 2x-{-Sy. 

6.  Divide  a2b2 — 2abzx  by  ab.     .     .     .     Ans.  ab — 2b2x. 

7.  Divide  \2albc— 9acx2+6ab2c  by  —  Sac. 

Ans.  —  4ab+Sx2—  2b2. 

8.  Divide  15a562c— 21aW  by  3a2&c.  Ans.  5afe— 7M& 

Note. — The  following  may  be  omitted  until  the  book  is  reviewed: 

9.  Divide  6(a+c)  +  9(a+.T)  by  3. 

Ans.  2(a+c)  +  3(a+je). 

10.  Divide  a2b(c-\-d)-\-ab2(c2—d)  by  ab. 

Ans.  a(c+tf)-}-&(c2— d). 

11.  Divide  ac(m-f-») — bc(m-\-ii)  by  »i+^  Ans.  ac — be. 

12.  Divide  (wi+w)(a54"^)2 +(»*+»)(* — "iff  Dy  »+«. 

Ans.  (xAryfArix—y)2. 

79.  To  explain  the  method  of  dividing  one  polynomial 
by  another,  we  will  first  find  the  product  of  two  factors, 
and  then  reverse  the  operation. 


Multiplication,  or  formation  of  a  product. 

2a2— ab 
a—b 


2d?—a2b 

—2a2b+ab2 

2a*  -3a26+a62 


Division,  or  decomposition  of  a  product. 

2a3— 3a26-fa62|a— b 
2a3— 2a2b  2a2— ab 

1st  Rem. 


-a2b+ab2 
-a2b+ab2 


2d  Rem. 


In  the  foregoing  illustration,  let  the  pupil  carefully  observe, 


DIVISION.  43 

1st.  In  the  multiplicand,  2a2 — ab,  and  the  multiplier,  a — 6,  a  is 
called  the  leading  Utter,  because  its  exponents  decrease  from  left  to 
right.  It  is  evident  that  a  will  be  the  leading  letter  in  the  prod- 
uct also. 

2d.  In  the  division  of  2a3— 3a2b-^ab2  by  a—b,  the  dividend 
being  the  product,  and  the  divisor  one  of  the  factors,  both  should  be 
arranged  with  reference  to  the  same  leading  letter,  in  order  that  the 
quotient,  or  remaining  factor  to  be  found,  may  have  the  same  order 
of  arrangement. 

3d.  If  we  divide  2a3  by  a,  the  result,  2a2,  will  be  the  term  of 
the  quotient  by  which  a — b  was  first  multiplied.  If  we  now  multi- 
ply a — 6  by  2a2,  and  subtract  the  product  from  the  dividend,  there 
will  remain  — a2b-\-ab2,  which  is  the  product  of  a — b  by  the  other 
term  of  the  quotient.  Dividing  — a2b  by  a,  we  find  this  unknown 
term.  Multiplying  a — b  by  it,  and  subtracting  the  product,  nothing 
remains. 

4th.  Had  there  been  a  second  remainder,  the  third  term  of  the 
quotient  would  have  been  found  in  the  same  manner,  and  so  on  for 
any  number  of  terms. 

5th.  The  divisor  is  placed  on  the  right  of  the  dividend  for  con- 
venience in  multiplying.     Hence, 


TO   DIVIDE    ONE    POLYNOMIAL    BY   ANOTHER, 

Rule. — 1.  Arrange  the  dividend  and  divisor  with  refer- 
ence to  the  leading  letter,  and  'place  the  divisor  on  the  right 
of  the  dividend. 

2.  Divide  the  frst  term  of  the  dividend  by  the  first  term 
of  the  divisor,  for  the  first  term  of  the  quotient.  Multiply 
the  divisor  by  this  term,  and  subtract  the  product  from  the 
dividend. 

3.  Divide  the  first  term  of  the  remainder  by  the  first  term 
of  the  divisor,  for  the  second  term  of  the  quotient.  Midtiply 
the  divisor  by  this  term,  and  subtract  the  product  from  the 
last  remainder. 

4.  Proceed  in  the  same  manner,  and  if  you  obtain  0  for  a 
remainder,  the  division  is  said  to  be  exact. 

Remarks. — 1.  Bring  down  no  more  terms  of  the  remainder,  at 
each  successive  subtraction,  than  arc  necessary. 


44 


RAY'S  ALGEBRA,  FIRST  BOOK. 


2.  It  is  well  to  perform  the  same  example  in  two  ways :  first,  by 
making  the  powers  of  the  letter  diminish  from  left  to  right;  and, 
secondly,  increase  from  left  to  right. 

3.  When  the  first  term  of  the  arranged  dividend,  or  of  any  re- 
mainder, is  not  exactly  divisible  by  the  first  term  of  the  arranged 
divisor,  the  exact  division  will  be  impossible. 


1.  Divide  6a2— 13ax+6x*  by  2a- 
6a2— 13ax+6x2\2a— Sx 
6a2 


Sx. 


-9ax 


—4;ax+6x2 
— 4ax-f6^2 


2.  Divide  x2 — y1  by  x — y. 

X2 — y2\x—y 

x2—xy      x^y  Quo. 

xy—y2 
xy—y2 


Sa—2x  Quotient. 


3.  Divide  a3+^  by  a-\-x. 

a3-f£3|a-f£ 
a3-f  a2x       a2—ax-\-x2  Quo. 

— a2x-\-xz 
— a2x— ax2 

ax2-j-xz 
ax2-\-x3 


4.  Divide  &a2x-{-5ax2-{-a?-\-x3  ^^  4a:c_|_a2_ra,2 

az-{-5a2x-{-5ax2-\-x?>\a2+4:ax-\-x2 
a3-)-4a2a:-|-arr2  a-\-x  Quotient. 

a2#-j-4a:z2-fa;3 

a2x-\-4ax2-^x- 


Note. — It  is  not  absolutely  necessary  to  arrange  the  dividend 
and  divisor  with  reference 'to  a  certain  letter;  it  should  be  done, 
however,  as  a  matter  of  convenience. 

In  the  above  example,  neither  divisor  nor  dividend  being  arranged 
with  reference  to  either  a  or  X,  we  arrange  them  with  reference 
to  a,  and  then  divide. 

Review. — 79.  What  is  meant  by  the  leading  letter?  What  is 
understood  by  arranging  the  dividend  and  divisor  with  reference  to 
a  certain  letter?  Explain  the  example  given  in  illustration  of 
division  of  polynomials. 

79.  Why  is  the  divisor  placed  on  the  right?  What  is  the  rule  for 
the  division  of  one  polynomial  by  another?  When  is  the  exact  divi- 
sion impossible? 


DIVISION. 


45 


5.  Divide  a2-\-a3 — 5a4+3a5  by  a — a2. 

a 

3a"> — 5a*+a3-f  a2| — a2-f-a 


Both  quantities  arranged  according  to 
the  ascending  powers  of  a. 


a2+a?—5a4-\-3ar>\a—a2 
a*-~  a* 


2a3— 5a4 
2a3  -2a4 


a+2a2— 3a3 
Quotient. 


-3a4+3a5 
-3a4+3a5 


Both  quantities  arranged  according  to 
the  descending  powers  of  a. 


3ar>— 3a4 


-2a4 + a3 
-2a4+2a3 


_3a3+2a2-j-a 
Quotient. 


— a3+a2 

— a3 -fa2 


It  will  be  seen  that  the  two  quotients  are  the  same,  but  differ- 
ently arranged.  If  preferred,  the  divisor  may  be  placed  on  the  left, 
instead  of  on  the  right,  of  the  dividend. 

G.  Divide  4a2— 8«x-\-4z!  by  2a— 2x.        Ans.  2a— 2x. 

7.  Divide  2xr-{-7xy-\-6y2  by  x-\-2y.  Ans.  2x-\-2>y. 

8.  Divide  Zmx+Snx+lQmn+l&n*  by  x-\-bn. 

Ans.  2wi+3«. 

9.  Divide  ar-f-2a-y+7/2  by  x-\-y Ans.  x-\-y. 

10.  Divide  8a4— 8x4  by  2a2— 2x\    .     .     Ans.  4a2-f  4a:2. 

11.  Divide  ac-\-~bc — ad — bd  by  a-\-b.  .     .       Ans.  c — d. 

12.  Divide  x3-\-y3-}-5xy2-\-5x2y  by  x2-\-4xy-\-y2. 

Ans.  ar-f-jf. 

13.  Divide  a3— 9a2-f-27a— 27  by  a—  3.  Ans.  a2—  6a+  9. 

14.  Divide  4a4— 5aV+x4  by  2a2—  3a:e+x2. 

Ans.  2a2+3a.T-f;c2. 

15.  Divide  x4 — y4,  by  a: — y.      .     .Ans.  a^-f-afy+a^+y*. 

16.  Divide  a3—h3  by  a2+aZ>-f-Z>2.      .      .     .       Ans.  a— b. 

17.  Divide  x3 — y3-\-3xy2 — Sx2y  by  cr — y. 

Ans.  .x2 — 2xy-\-y2. 

18.  Divide  4.c4— 04  by  2x— 4.  Ans.  2.c3+4.x2-f-8^+16. 

19.  Divide  a5— 5a4.x-fl0a3.c2— 10a2.r3+5ax4— a;5  by  a2 
— 2aa:+rc2.  Ans.  a3 — 3a2.-E-j-3acc2 — x3. 

20.  Divide  4a6— 25aV+20r^— 4.r6  by  2a3— 5a*2+2ar\ 

Ans.  2a3-j-5aa2— 2a3. 

21.  Divide  ?/3+l  byy-fl Ans.  y2 — y-\-l. 


40  RAYS  ALGEBRA,  FIRST  BOOK. 

22.  Divide    6a4-f-4a3a>— 9a2x2— 3«^-f~2.T4   by   2a*+2ax 
—x'2.  Ans.  3a2 — ax — 2a2. 

23.  Divide  3a4— 8a2Z>2+3a2c2+5Z/4— 3^2c2  by  a2— h\ 

Ans.  3a2—  5Z>2+3c2. 

24.  Divide  x*—3xy-\-3xy—f  by  a3— 3.x2?/+ 3;zy2— y». 

Ans.  xd-\-3x2y+3xf+f. 

MISCELLANEOUS     EXERCISES. 

1.  3a-f-5z  —  dc+ld+ba—  Sx— 3d  —  (4a+2x-8c-f 
4d)=what?  Ans.  4a— c. 

2.  a+6— (2a— 3o)— (5a+7o)— (— 13a+2&)=what  ? 

Ans.  7a — bb. 

3.  (a+&)(a+&)  +  (a— 6)(a— 6)=what?  Ans.  2a*+26». 

4.  (a2+a4-f  a6)(a2— 1)— (a44-a)(a4— a)=what?  Ans.  0. 

5.  (a3-fa2Z>— a&2— 63)^(a— 6)—  (a— &)(a— 6)=what? 

Ans.  4a6. 


II.    ALGEBRAIC    THEOREMS, 

DERIVED  FROM   MULTIPLICATION  AND  DIVISION. 

80.  If  we  square  a-j-6,  that  is,  multiply  a-\-b  by  itself, 
the  product  will  be  a2+2ab-\-b2. 

Thus:  ct-j-6 


a2-}-a6 

a2_|_2a6-|-62 

But  a-\-b  is  the  sum  of  the  quantities,  a  and  b.     Hence, 

Theorem  I. —  Hie  square  of  the  sum  of  two  quantities  is 
equal  to  the  square  of  the  first,  plus  twice  the  product  of  the 
first  by  the  second,  plus  the  square  of  the  second. 


ALGEBRAIC  THEOREMS.  47 

Note. — Let  the  pupil  apply  tlic  theorem  by  writing  the  following 
examples,  enunciated  thus:  What  is  the  square  of  2-J-3? 

1.  (2-f-3)2=4+12+9=25. 

2.  (2a-\-hy=4:a2+4:ab-\-L\ 

3.  (2x-\-3>/y=4x2+12ccy+9f. 

4.  lab+cciy=dzh2-)-2ahcd+c'id\ 

5.  (x!-{-xyy=xi-{-2xii/-\-xy. 

6.  (2a2+3az)2==4a4+12a3;r+9aV. 

81.  If  we  square  a — 6,  that  is,  multiply  a — b  by  itself, 
the  product  will  be  a2 — 2ab-\-b2. 

Thus:  a—b 
a — b 


a2 — ab 
—ab+b2 

a2—2ab-{-b2 

But  a — b  is  the  difference  of  the  quantities,  a  and  b. 
Hence, 

Theorem  II. — The  square  of  the  difference  of  two  quan- 
tities is  equal  to  the  square  of  the  first,  minus  twice  the  prod- 
uct of  the  first  by  the  second,  plus  the  square  of  the  second. 

1.  (5— 4)2=25-404-16=l. 

2.  (2a— by=4d2— 4ab+b\ 

3.  (Sx—2I/y=9x2— \2xy+±y\ 

4.  {x1—fy=x'—2xhf^-y\ 

5.  (ax— x2y=a*x2— 2axi+xi. 

6.  (5a2— fc2)2=25a4— 10a262+&*. 

82.  If  we  multiply  a-\-b  by  a — b,  the  product  will  be 
a2— b\ 

Thus:  a-\-b 
a—b 

ctf+ab 
—ab—b2 

a2—b2~ 


48  RAY'S  ALGEBRA,  FIRST  BOOK. 

But  a-\-b  represents  the  sum  of  two  quantities,  and 
a — b,  their  difference.     Hence, 

Theorem  III. — The  product  of  the  sum  and  difference  of 
two  quantities  is  equal  to  the  difference  of  their  squares. 

1.  (5+3)(5— 3)=25— 9=16=8x2. 

2.  (2a+6)(2a— h)=4a2— b\ 

3.  (2*+3j/)(2tf— %)=4**—  %2. 

4.  (5a+46)(5a~4&)=25«2— 1662. 

6.   (2a»i+3&»)(2ani—  3Zm)=4a2m2—  9h2n\ 

8«5.  If  we  divide  a3  by  a5,  observing  the  rule  for  the 

exponents,  we  have  _=a3_5=a~2.    But,  Art.  127,  ~— -. 
ab  a5     a1 

CLm                         am         1                                       1 
So,  — =am-n:   and    _-=_—.     Hence,  0"*-*= ,. 

'  an  '  an      an~m  a"-"1 

a63      a 

=,    Q  ,  etc.     Hence, 

c      6~3c  ' 

Theorem  IV. — 1.  The  reciprocal  of  a  quantity  is  equal  to 
the  same  quantity  with  the  sign  of  its  exponent  changed. 

2.  Any  quantity  may  be  transferred  from  one  term  of  a 
fraction  to  the  other,  if  the  sign  of  the  exponent  be  changed. 

Thus:     ....     j-=ab-i= — -= — -; 

<*W_  _,_        I .  __c-*d-* 

~c2(P~       °        '>~a-ib-^df~a~'^b-^ 

a2 


84.  By  the  rule  for  the  exponents,  Art.  74,  — =a2  2=:a° ; 

a2 

a2 
but  since  any  quantity  is  contained  in  itself  once, =1. 


a,  a 

Similarly,  — =am-TO=a0 ;    but    -^=1;   therefore,  a°=l,   since 

am 
each  is  equal  to  — — .     Hence, 


ALGEBRAIC  THEOREMS.  49 

Theorem  V. — Any  quantity  whose  exponent  is  0  is  equal 
to  unity. 

85.  If  we  divide  a2 — b2  by  a — 6,  the  quotient  will  be 
a-\-b.  If  we  divide  a3 — b3  by  a — fe,  the  quotient  will 
be  a*j-ab-ftr*. 

In  the  same  manner,  we  should  find,  by  trial,  that  the 
quotients  obtained  by  dividing  the  difference  of  the  same 
powers  of  two  quantities  by  the  difference  of  those  quan- 
tities, follow  a  simple  law. 

Thus:     (a2— 62)--(a— b)=a-\-b. 

(a3— ^--(a— 6)=a2-fa&-}-&2. 
(a4_&4)^  (a__6)=a3-|-  a?b+ aW-\-  6*. 
(a5_65)V(a— 6)=a4-fa364-a262-4-a63-|-64. 

So,  (a5— l)-(a— l)=a4+«3+«2-F«+l. 
And  (l-6(i)-(l-6)=l-r-6-r-62-r-63-|-64. 

The  exponent  of  the  first  letter  decreases  by  unity,  while 
that  of  the  second  increases  by  unity.     Hence,  we  have 

Theorem  VI. —  The  difference  of  the  same  powers  of  two 
quantities  is  always  divisible  by  the  difference  of  the  quantities. 

86.  The  two  following  theorems  may  also  be  readily 
shown  to  be  true  by  trial : 

Theorem  VII. —  The  difference  of  the  even  powers  of  two 
quantities  of  the  same  degree,  is  always  divisible  by  the  sum 
of  the  quantities. 

Thus:     (a2— 62)-~(«-f-6)r=a— 6. 

(a4_54)_^a_|_&)=a3_a2&_j_ot&2_&3> 

(af>_&f>)_^(a_j_&)=:a5— a*b-\-aW— a2tfl+ab*— b\ 

So,  (a6— l)-r-(a-|-l)=a5— a4-fa3— a2+a— 1. 
And  (1—66)--  (i_f_&)=i_ &_j_&2_63_|_&4_ #> 

Review. — 80.  To  what  is  the  square  of  the  sum  of  two  quantities 
equal?  81.  Of  the  difference  of  two  quantities?  82.  The  product 
of  the  sum  and  difference  of  two  quantities? 

83.  How  may  the  reciprocal  of  any  quantity  be  expressed?  How 
may  any  quantity  be  transferred  from  one  term  of  a  fraction  to  the 
other?     In  what  other  form  may  am  be  written?     a—m? 

84.  What  is  the  value  of  any  quantity  whose  exponent  is  zero? 

1st  Bk.  5 


50  RAY'S  ALGEBRA,  FIRST  BOOK. 

Theorem  VIII.  —  The  sum  of  the  odd  powers  of  two  quan- 
tities of  the  same  degree,  is  always  divisible  by  the  sum  of  the 
quantities. 

Thus:  (a3-f-63)-r-(a4-6)=a2— a6-f-62. 

(a5^.63)_^(a_|_6)=a4— a-'b-{-a2b2—ab^bK 

(a7^67^(a_|_6)^ar,_a5^_|_a462_a3^_|_a254_a^_j_&f,> 

So,.(a7^l)^(a-f-l)=a6— a5-i-a4— a3-}-a2— a-j-1. 

And  (14-a7)_f.(l^_a)=l_a+a2_a3^_a4_a5_j_af,> 

Note. — For  a  complete  demonstration  of  Theorems  vi,  vii,  and 
viii,  see  Ray's  Algebra,  Second  Book,  Arts.  83,  84,  85,  and  86. 


FACTORING. 

FACTORS,   AND  DIVISORS  OF  ALGEBRAIC  QUANTITIES. 

ST.  A  Divisor  or  Measure  of  a  quantity  is  any  quan- 
tity that  divides  it  -without  a  remainder. 

Thus,  2  is  a  divisor  or  measure  of  6,  and  a2  of  a2x. 

88.  A  Prime  Number  is  one  which  has  no  divisors  ex- 
cept itself  and  unity. 

A  Composite  Number  is  one  which  has  one  or  more 
divisors  besides  itself  and  unity.     Hence, 

All  numbers  are  either  prime  or  composite  ;  and  every 
composite  number  is  the  product  of  two  or  more  prime 
numbers. 

The  prime  numbers  are  1,  2,  3,  5,  7,  11,  13,  17,  etc. 

The  composite  numbers  are  4,  6,  8,  9,  10,  12,  14,  15, 
16,  etc. 

TO  RESOLVE  A  COMPOSITE  NUMBER  INTO  ITS  PRIME  FACTORS, 

Rule. — 1.  Divide  by  any  prime  number  that  will  exactly 
divide  it;  divide  the  quotient  again  in  the  same  manner,  and 
so  continue. 

Review. — 85.  By  what  is  the  difference  of  the  same  powers  of  two 
quantities  always  divisible?  86.  The  difference  of  the  even  powers 
of  the  same  degree?     The  sum  of  the  odd  powers? 


FACTORING.  51 

2.  Tiie  last  quotient  and  the  several  divisors  will  be  the 
prime  factors* 

Remark. — It  is  most  convenient  to  divide  first  by  the  smallest 
prime  number  that  is  a  factor. — See  Ray's  Arithmetic,  Third  Book, 
Factoring. 

1.  What  are  the  prime  factors  of  105  ?         Ans.  3,  5,  7. 

2.  What  are  the  prime  factors  of  210  ?    Ans.  2,  3,  5,  7. 

3.  Resolve  4290  into  its  prime  factors. 

Ans.  2,  3,  5,  11,  13. 

80.  A  Prime  Quantity,   in  Algebra,  is  one  which  is 
exactly  divisible  only  by  itself  and  by  unity. 
Thus,  a,  b,  and  b-\-c  are  prime  quantities. 

90.  Two  quantities  are  prime  to  each  other  when  no 
quantity  except  unity  will  exactly  divide  both. 

Thus,  ab  and  cd  are  prime  to  each  other. 

91.  A  Composite  Quantity  is  one  which  is  the  product 
of  two  or  more  factors,  neither  of  which  is  unity. 

Thus,  ax  is  a  composite  quantity,  the  factors  being  a  and  x. 

92.  To  separate  a  monomial  into  its  prime  factors, 

Rule. — Resolve  the  coefficient  into  its  prime  factors.  To 
these  annex  the  literal  factors. 

Find  the  prime  factors  of  the  following  monomials : 

1.  15a2bc Ans.  Sx&.a.a.b.c. 

2.  21ab2d Ans.  3x*?.a.b.b.d. 

3.  35abc*x Ans.  5x*7 .a.b.c.c.x. 

Review. — 87.  What  is  the  divisor  of  a  quantity?  88.  A  prime 
number?  A  composite  number?  Name  several  of  the  prime  num- 
bers, beginning  with  unity.  Composite  numbers,  beginning  with  4. 
Rule  for  resolving  any  composite  number  into  its  prime  factors? 

89.  What  is  a  prime  quantity?  Example.  90.  When  are  two 
quantities  prime  to  each  other?     Example. 

91.  What  is  a  composite  quantity?  Example.  92.  Rule  for  sepa- 
rating a  monomial  into  its  prime  factors? 


52  RAY  S  ALGEBRA,  FIRST  BOOK. 

93.  To  separate  a  polynomial  into  its  factors  when  one 
of  them  is  a  monomial  and  the  other  a  polynomial. 

Rule. — Divide  the  given  quantity  by  the  greatest  monomial 
that  will  exactly  divide  each  of  its  terms;  the  divisor  will  be 
one  factor  and  the  quotient  the  other. 

Separate  the  following  expressions  into  factors : 

1.  x+ax Ans.  ar(l-fa). 

2.  am-\-ac Ans.  a(m-f  c). 

3.  btf+bcd Ans.  bc(e+d). 

4.  4x2+6xy Ans.  2x(2x+3y). 

5.  6ax2y+9bxy2— \2vx2y.  .      .Ans.  Sxy(2ax+  Shy— icx). 

6.  hax2 — <&baxiy-\-bd2xzy.  .     .   Ans.  5ax2(l — *7xy-\-axy). 

7.  a3cm2-\-a2c2m2 — a2cmz.    .     .     .     Ans.  a2cm2(a-\-c — m). 

94.  To  separate  a  quantity  which  is  the  product  of  two 
or  more  polynomials  into  its  prime  factors. 

No  general  rule  can  be  given  for  this  case.  When  the  given 
quantity  does  not  consist  of  more  than  three  terms,  it  may  gener- 
ally be  accomplished  by  reversing  some  one  of  the  preceding 
theorems. 

1st.  For  a  trinomial,  whose  extremes  are  squares  and  positive,  and 
whose  middle  term  is  twice  the  product  of  the  square  roots  of  the 
extremes,  reverse  Arts.  80  and  81. 

Thus:   a2+2a6+62=(a+6)(a+6). 
a2— 2a6-f62=(a— b)(a— 6), 

2d.  For  a  binomial,  which  is  the  difference  of  two  squares, 
verse  Art.  82. 
Thus:   a2—b2=(a+b)(a—b). 

3d.  For  the  difference  of  the  same  powers  of  two  quantities,  re- 
verse Art.  85. 
Thus:   a3— &=(a— b)(a2+ab+b2). 

a5~6-r»=(a— 6)(a44-a36-fa262+a63-f&4). 

4th.  The  difference  of  the  even  powers  of  two  quantities,  higher 
than  the  second  degree,  may  be  separated  thus : 

a*— b*=(a2+b2)(a2—b2)=(a2+b2)(a+b)(a—b). 
a&—b&  =  (a3+63)  (a3— 63)  =  (a+6)  (a2— ab+b2)  (a—b)  (a2+ 
ab+b2). 


FACTORING. 


53 


5th.  For  the  sum  of  the  odd  powers  of  two  quantities,  sec  Art.  86. 
Thus:      a3  +  &3==(a  +  6)(a2— «&-f&2). 

a7+67==(o+6)(a6— abb+aAb2— a363+a264— a65+6fi). 

Separate  the  following  into  their  simplest  factors : 

1.  *s-j-2ary+jf.  7.  9m2— 16>i2. 

2.  9e*2-|-12a&-f  4F. 

3.  4+12^+9^. 

4.  m2 — 2wi»-f-»8. 

5.  4z2— 20z.:4-25z2. 


G. 


1. 


O.    iC 

9.  y- 

10.  ^—1. 

11.  8a3— 2lb\ 

12.  a5-f7A 


ANSWERS. 


2.  (3a+26)(3a-f  2&). 

3.  (2+3*)(2+3s). 

4.  (in — n)  (wi — w). 

5.  (2z—  fc)(2*Mb). 

7.  (3w+4«)(3m— 4»). 


8.  (a;2-f £-)(>  — i2)=(:r2-f- 

io.  (x— ix^+jp^i)'. 

11.  (2a— 3^)  (4a2-j-G^-f 

12.  (a-fZ>)(a4— a3Z>-f  «"&2 


95.  To  separate  a  quadratic  trinomial  into  its  factors. 

A  Quadratic  Trinomial  is  of  the  form  a?-\-ax-\-b,  in 
which  the  signs  of  the  second  and  third  terms  may  be 
either  plus  or  minus. 

Such  quantities  may  be  resolved  into  two  binomial  fac- 
tors by  inspection. 

Thus:  X2— 5x-\- 6  will  evidently  be  the  product  of  X—2  by  X— 3. 

Review. — 93.  Rule  for  separating  a  polynomial  into  its  prime 
factors,  when  one  of  them  is  a  monomial  and  the  other  a  polynomial. 
94.  When  can  a  trinomial  be  separated  into  two  binomial  factors? 

94.  What  are  the  factors  of  m2+2mn+n2?    Of  c2— 2co7+d2?    When 
can  a  binomial  be  separated  into  two  binomial  factors?     What,  the 
factors  of  x2- 
a2—b21     Of  a3— 63? 
a4_£4?     Ofa,;— i(i? 


,2?     Of  9a2— lti&2?     What  is  one  of  the  factors  of 
Of  x4 — y*  ?     What  are  two  of  the  factors  of 


54  RAY'S  ALGEBRA,  FIRST  BOOK. 

Decompose  each  of  the  following  trinomials  into  two 
binomial  factors : 

1.  &-\- O24-6 Ans.  (x-\- 2)(a?+3) 

2.  a2+7a-f-12 Ans.  (a+3)(a+4) 

3.  a:2— 5*-f  6 ' .     .     .  Ans.    (ar— 2)(*— 3) 

4.  a?2+a?— 6 Ans.  (a?+  3)(a?— 2) 

5.  ap'-f-ap — 2 Ans.  (#+2)(a:— 1) 

6.  a?'— 13ar+40 Ans.  (x—  $)(x—  5) 

7.  *2— 7z— 8 Ans.  (a?_8)(a?+l) 

8.  *»—  ar—  30 Ans.  (*_6)(>+5) 

We  may  often  separate  other  trinomials  into  factors  by  first  tak 
ing  out  the  monomial  factor  common  to  each  term. 

Thus,  5ax2— lOax —40a=5a(x2— 2x— 8)=ba(x— 4)  (ar-f  2). 

9.  3z2-f-12*— 15 Ans.  3(x+5)(x— 1). 

10.  2abx2— 14abx-~ GOab.     .     Ans.  2a6(a>— 10)(a;+3). 

11.  2x3— 4x2—30z Ans.  2a<a5— 5)(a;+3). 

06.  The  principal  use  of  factoring  is  to  shorten  alge- 
braic operations  by  canceling  common  factors. 

Whenever  there  is  an  opportunity  of  doing  this,  the 
operations  to  be  performed  should  be  merely  indicated  as 
in  the  following  examples : 

1.  Multiply  a — b  by  x2-{-2xy-\-y2,  and  divide  the  prod- 
uct by  x-\-y. 

(a-b)(x2+2xy+y2)Ja-b)(x^j)(xJry)__ 

x-^y  x-\-y  *         "  ~*y' 

=ax-{-ay — bx — by. 

2.  Multiply  x — 3  by  x2 — 1,  and  divide  the  product 
by  x — 1,  by  factoring.  Ans.  x2 — 2x — 3. 

3.  Divide  ^-f-l  by  s+l>  and  multiply  the  quotient 
by  z2 — 1,  by  factoring.  Ans.  z* — z3-\-z — 1. 

4.  Multiply  x2—  5.r-f  6  by  x2—  7^+12,  and  divide  the 
quotient  by  x2 — 6.T+9,  by  factoring.     Ans.  (x — 2) (a; — 4). 

Review. — 94.  What  is  one  of  the  factors  of  az-\-l?l  What  is  one 
of  the  factors  of  x^-\-yr°  ?     95.  What  is  a  quadratic  trinomial  ? 


GREATEST  COMMON  DIVISOR.  55 


GREATEST    COMMON    DIVISOR. 

9*7.  A  Common  Divisor,  or  Common  Measure,  is  any 

quantity  that  will  exactly  divide  two  or  more  quantities. 

Thus,  2  is  a  common  divisor  of  8  and  12  ;  and  a  is  a 
common  divisor  of  ab  and  a2x. 

Remark. — Two  quantities  may  sometimes  have  more  than  one 
common  divisor.    Thus,  8  and  12  have  two  common  divisors,  2  and  4. 

98.  The  Greatest  Common  Divisor,  or  Greatest  Com- 
mon Measure,  of  two  or  more  quantities,  is  the  greatest 
quantity  that  will  exactly  divide  each  of  them. 

Thus,  the  greatest  common  divisor  of  4a2xy  and  6a?x2y2 
is  2a2xy. 

99.  Quantities  that  have  a  common  divisor  are  said  to 
he  commensurable;  and  those  that  have  no  common  divisor 
incommensurable. 

Note. — G.C.D.  stands  for  greatest  common  divisor. 

100.  To  find  the  G.  CD.  of  two  or  more  monomials. 

1.  Let  it  be  required  to  find  the  G.C.D.  of  Qab  and 
15a2c. 

By  separating  each  quantity  into  its  prime  factors,  we  have 
6a6z=2x3a6,  15a2c=3x5aac. 

Here,  3  and  d  are  the  only  factors  common  to  both  terms ;  hence, 
both  can  be  divided  either  by  3  or  a,  or  by  their  product  3a,  and  by 
no  other  quantity  ;  consequently,  3a  is  their  G.C.D.     Hence, 

TO  FIND  THE  GREATEST  COMMON  DIVISOR  OF  TWO  OR  MORE 
MONOMIALS, 

Rule. — 1.   Resolve  the  quantities  into  their  prime  factors. 

2.  Take  the  product  of  those  factors  that  are  common  to 
each  of  the  terms  for  the  greatest  common  divisor. 

Note. — The  G.C.D.  of  the  literal  parts  will  be  the  highest  power 
of  each  letter  which  is  common  to  all  the  quantities. 


56  RATS  ALGEBRA,  FIRST  BOOK. 

2.  Find  the  G.C.D.  of  4a2.x3,  6a?x2,  and  KW.t. 

4a2rc3=2x2a2£3  Here,  we  see,  that  2.  a2,  and  x  are  the  only 

6a^x2=2ySa^x2         factors  common  to  all  the  quantities.    Hence, 
10a*x  =2x^a4x  2a'2%  is  the  G.C.D. 

Find  the  G.C.D.  of  the  following  quantities : 

3.  4<x2x2,  and  10ax3 Ans.  2ax2. 

4.  4a36V/,  and  Sa6x2y2 Ans.  4aV/. 

5.  SaxYsP,  12x>z%  and  24a3xV2.       .      .     .  Ans.  4xh\ 

6.  6a2xy27  12ay^,   9a5^y,  and  2^a3fz.    .  Ans.  3a2/. 

101.  To  find  the  G.C.D.  of  two  polynomials. 

Previous  to  the  investigation  of  this  subject,  it  will  be 
necessary  to  state  the  following  propositions : 

Proposition  I. — A  measure  or  divisor  of  a  quantity  is  also 
a  measure  of  any  number  of  times  that  quantity. 

Thus,  if  3  is  a  measure  of  6,  it  is  a  measure  of  2x6  or  12,  of 
3X6  or  18,  etc. 

Proposition  II. — A  common  measure  of  two  quantities  is 
also  a  measure  of  their  sum. 

Thus,  if  3  is  a  common  measure  of  27  and  18,  it  is  also  a  meas- 
ure of  27-|-18  or  45,  since  27-J-18  divided  by  3=9-}-6=15,  and 
45—3=15.    i 

Proposition  III. — A  common  measure  of  two  quantities  is 
also  a  measure  of  their  difference. 

Thus,  if  3  measures  27  and  18,  it  will  also  measure  27—18  or  9, 
since  27—18  divided  by  3=9—6=3,  and  9-=-3=3.  It  is  also  evi- 
dent that  the  greatest  common  measure  of  27  and  18  is  also  the 
greatest  common  measure  of  27,  18,  and  9;  that  is,  of  the  quantities 
themselves  and  their  difference. 

102.  Let  it  be  required  to  find  the  G.C.D.  of  36 
and  116. 

If  we  divide  116  by  36,  and  there  is  no  remainder,  36  is  evidently 
the  G.C.D.,  since  it  can  have  no  divisor  greater  than  itself.     Divid- 


GREATEST  COMMON  DIVISOR.  57 

ing  116  by  36,  we  find  the  quotient  is  3,  and  36)116(3 

the  remainder  is  8,  which  is   necessarily  less  108 

than  either  of  the  quantities  116  and  36,  and  8^36(4 

by  Prop.  3,  Art.  101,  is  exactly  divisible  by  32 

their  G.C.D.;  hence,  their  G.C.D.  must  divide  — 

116,  36,  and  8.  4)8(2 

8 
Now,  if  8  will  exactly  divide  36,  it  will  also  

exactly  divide  116.  since  it  must  divide  3x36  0 

or  108,  and  also  108-}-8  or  116.  Prop's  1  and  2, 

Art.  101,  and  will  be  the  G.C.D.  sought.     As  it  does  not,  it  remains 

to  find  the  G.C.D.  of  36  and  8. 

Dividing  36  by  8,  the  quotient  is  4,  with  a  remainder  4.  Reason- 
ing as  before,  the  G.C.D.  must  divide  36,  8,  and  4,  and  can  not  be 
greater  than  4. 

Now,  if  4  will  exactly  divide  8,  it  will  also  exactly  divide  36, 
since  36=4x8-f-4,  and  will  be  the  G.C.D.  sought.  By  trial,  we  find 
that  4  will  divide  8,  leaving  no  remainder. 

Since,  then,  the  G.C.D.  of  116  and  36  must  be  the  G.C.D.  of 
116,  36,  8,  and  4,  and  since  4  has  been  shown  to  be  a  measure 
of  8,  36,  and  116,  and  since  4  is  the  greatest  measure  of  itself,  it 
follows  that  4  is  the  number  sought. 

103.  Suppose,  now,  that  it  is  required  to  find  the 
G.C.D.  of  two  polynomials,  A  and  B,  of  which  A  is  the 
greater. 

Let  the  successive  quotients  and  remainders      B)A(Q 
be  represented  by  Q,  0/,  Q",  etc.?  and  R,  R',       BQ 

R//> etc'  ~R)B(Cr 

By  the  same  process  of  reasoning  as  in  the  -dq/ 

example  above,  it  may  be  shown  that  R'  being  

the  greatest  measure  of  itself,  and  also  a  meas-  >    v^ 

P/q// 
ure  of  R,  is  the  greatest  common  measure  of  _L 

RXQ/-f-R/  which  is  equal  to  B,  and  of  BXQ-f-R,  0 

which  is  equal  to  A,  or  that  it  is  the  G.C.D.  of 

A  and  B. 

Review. — 95.  How  can  a  quadratic  trinomial  be  separated  into 
binomial  factors?  96.  What  is  the  principal  use  of  factoring? 
97.  What  is  a  common  divisor  of  two  or  more  quantities?    Example. 

98.  What  is  the  G.C.D.  of  two  quantities?  Example.  100.  How 
find  the  G.C.D.  of  two  or  more  monomials?  101.  State  the  three 
propositions  in  Art.  101,  and  illustrate  them. 


58  RAYS  ALGEBRA,  FIRST  BOOK. 

103.  To  whatever  extent  the  division  is  carried,  the 
process  of  reasoning  is  the  same,  and  the  last  divisor  will 
be  the  G.C.D.  When  this  last  divisor  is  unity,  or  does  not 
contain  the  letter  of  arrangement,  there  is  no  common 
divisor  to  the  quantities. 

104.  If  one  of  the  quantities  contain  a  factor  not  found 
in  the  other,  it  may  be  canceled  without  affecting  the  com 
mon  divisor.      (See  Exam.  3.) 

If  both  quantities  contain  a  common  factor,  it  may  be 

set  aside  as  a  factor  of  the  common  divisor ;  and  we  may 

proceed  to  find  the  G.C.D.  of  the  other  factors  of  the  given 

quantities.      (See  Exam.  2.) 

2a  bx 
Thus,  in  the  fraction  ^—s-.  the  G.C.D.  of  the  two  terms  is  evi- 
Sabc 

dently  ab.     If  we  cancel  2  or  x_  in  the  numerator,  or  3  or  c  in  the 

denominator,  ab  is  still  the  common  divisor. 

Again,  in  the  fraction    — — ,  a  is  a  part  of  the  common  divisor. 

276 
Setting   this  aside,   and  finding  the  common  divisor  of  —^-,  which 

aX 

is  9,  we  have  for  the  G.C.D.  of  the  original  fraction  ayS^a. 

105.  We  may  multiply  either  quantity  by  a  factor  not 
found  in  the  other,  without  affecting  the  G.C.D. 

m,        .  2abx  , 

Thus,  in  the  fraction  - — — ,  wJiose  G.C.D.  is  ab,  if  we  multiply 
3abc  * 

the   dividend   by  4,    a   factor   not   found  in    the   divisor,  we  have 

8abx 

n    .    ,  of  which  the  common  divisor  is  still  ab. 

Sabc' 

In  like  manner,  if  we  multiply  the  divisor  by  any  factor  not 

found  in  the  dividend,  the  common  divisor  will  remain  the  same. 

If,  however,  we  multiply  the  numerator  by  3,  which  is  a  factor 

of  the  denominator,  tho  result  is  - — p- 1  of  which  the  G.C.D.  is  Sab. 

3a6c 

and  not  ab  as  before. 

Hence,   the   G.C.D.  may  be  changed  by  multiplying  one  of  the 
quantities  by  a  factor  of  the  other. 

106.  In  the  general  demonstration,  Art's  101,  102,  it 
has  been  shown  that  the  G.C.D.  of  two  quantities  exactly 


GREATEST  COMMON  DIVISOR.  59 

divides  each  of  the  successive  remainders.  Hence,  the 
preceding  principles  likewise  apply  to  the  successive  re- 
mainders. 

1.  Find  the  G.C.D.  of  x5— f  and  x*—x2y\ 

Here,  the  second  quantity  contains  x2  as  a  factor,  but  it  is  not  a 
factor  of  the  first;  we  may,  therefore,  cancel  it,  and  the  second 
quantity  becomes  X2—y2.     Divide  the  first  by  it. 

After  dividing,  we  find  that  y2  is  a  factor  of  x?>— y3  \x2—y2 
the  remainder,  but  not  of  x2—y2,  the  dividend,  x3—xy2  (x 
Hence,  Xy2—yz 

By  canceling  it,  the  divisor  becomes  x— y;      Qr      tX—y)y2 
then,  dividing  by  this,  we  find  there  is  no  re- 
mainder; therefore,  x—y  is  the  G.C.D. 


x2- 

-y2 

-xy 

\x- 

-y 

X2- 

{x+y 

xy- 

-y2 

xy- 

-y2 

■aV 

X3- 

-a2x 

\X2- 

a2 

X3- 

(X 

2.  Find  the  G.C.D.  of  x6-\-a3x3  and  x*- 

The  factor  x2  is  common  to  both  these  quan- 
tities; it  therefore  forms  part  of  the  G.C.D.,  and 
may  be  taken  out  and  reserved.  Doing  this,  the 
quantities  become  Xx-\-azx  and  X2 — a2. 

The  first  quantity  still  contains  a  common 
factor,  X,  which  the  latter  does  not;  canceling      x2     a2    \x-\-d 

this,  it  becomes  x3-\-ds.    Then,  proceeding  as      x2-Lax  (x < 

in   the   first   example,    we   find   the    G.C.D.   is     „ 

{  T   ;  -ax— a2 


(x-\-a)a2 


3.   Find  the  G.C.D.  of  5a5-f-10a4o:-f  ha3x2  and  a3x-\- 2a2x2 
-f2ax3+^. 

Here,  5a3  is  a  factor  of  the     as+2a2x+2ax2+x:''\a2-}-2ax+x2 
first   quantity   only,   and  x  of      «3-f  2a2xArax2  (a 

the  second  only.  ax2-\-xz 

Suppressing    these    factors,  or    (a4-x)x2 

and  proceeding  as  in  the  pre- 
vious examples,  we  find  a+x  is      a2-\-2ax+x2     \a~\-x 
the  G.C.D.  a2+ax  («+aT 

ax+x2 
ax-{-x2 


60 


RAY'S  ALGEBRA,  FIRST  BOOK. 


4.   Find   the  G.C.D.  of  2a4- 
2aV— 3x*. 


-a'x- 


■6x*  and  4tab-\-6a3x2- 


In  solving  this  ex- 
ample, there  are  two 
instances  in  which  it 
is  necessary  to  multi- 
ply the  dividend,  in 
order  that  the  coeffi- 
cient of  the  first  term 
may  be  exactly  divis- 
ible by  the  divisor. 
See  Art.  105. 

The  G.C.D.  is  found 
to  be  2a2-{-3z2. 


Aw+Ga-'x2— 2a2z3— 3tf>     12a*— a2:c2— 6ar* 
±a:>—2a?>x2—l2ax*  (2a 


8a3a2— 2a2ic3-j-12a^_3a;5 
or,  (8a^—2a2x-\-12ax2—3x:i)x2 

2a^—a2x2—6xi 
4 


8a*—4a2x2—24xi  |  8a3— 2a2x-]-l  2ax2—3x* 
8a*— 2a3#-fl 2a2x2—3ax:i  (a 

2a%— 1 6a2x2-\-3ax*— 24z  * 
4 


8a3rc— 64a2x2-\-l  2ax3— 96a: ;  (x 
8a*x—  2a2x2-\-\2atf>—  3x* 


-62a2x2- 


-93x* 


or,  — 31:r2(2a2-|-3z2) 


8a3— 2a2:c +12ax2—3x"  1 2a2-| 
8a3  -j-12a#2  ~0a- 


■3x2 


—2a2x- 
—2a2x- 


-3z3 
-3z3 


Hence, 


TO    FIND   THE    GREATEST    COMMON    DIVISOR   OF  TWO   POLY- 
NOMIALS, 

Rule. — 1.  Divide  the  greater  polynomial  by  the  less,  and 
if  there  is  no  remainder,  the  less  quantity  will  be  the  divisor 
sought. 

Review. — 102.  Explain  the  rule  for  finding  the  G.C.D.  of  two 
numbers,  as  illustrated.  103.  When  do  we  conclude  that  there  is  no 
common  divisor  to  two  quantities? 

104.  How  is  the  common  divisor  of  two  quantities  affected  by 
canceling  a  factor  in  one  of  them,  not  found  in  the  other?  When 
both  quantities  contain  a  common  factor,  how  may  it  be  treated? 

105.  How  is  the  G.C.D.  of  two  quantities  affected  by  multiplying 
either  of  them  by  a  factor  not  found  in  the  other? 

106.  Rule  for  finding  the  G.C.D.  of  two  polynomials?  How  find 
the  G.C.D.  of  three  or  more  quantities? 


GREATEST  COMMON  DIVISOR.  61 

2.  If  there  is  a  remainder,  divide  the  first  divisor  by  it, 
and  continue  to  divide  the  last  divisor  by  the  last  remainder, 
until  a  divisor  is  obtained  which  leaves  no  remainder^'  this 
will  be  the  greatest  common  divisor  of  the  two  given  poly- 
nomials. 

Notes. — 1.  When  the  highest  power  of  the  leading  letter  is  the 
same  in  both,  it  is  immaterial  which  of  the  quantities  is  made  the 
dividend. 

2.  If  both  quantities  contain  a  common  factor,  let  it  be  set  aside, 
as  forming  a  factor  of  the  common  divisor,  and  proceed  to  find  the 
G.C.D.  of  the  remaining  factors,  as  in  Ex.  2. 

3.  If  either  quantity  contain  a  factor  not  found  in  the  other,  it 
may  be  canceled  before  commencing  the  operation,  as  in  Ex.  3.  See 
Art.  104. 

4.  When  necessary,  the  dividend  may  be  multiplied  by  any  quan- 
tity which  will  render  the  first  term  divisible  by  the  divisor.  See 
Art.  105. 

5.  If,  in  any  case,  the  remainder  does  not  contain  the  leading 
letter,  that  is,  if  it  is  independent  of  that  letter,  there  is  no  common 
divisor. 

6.  To  find  the  G.C.D.  of  three  or  more  quantities,  first  find  the 
G.C.D.  of  two  of  them;  then,  of  that  divisor  and  one  of  the  other 
quantities,  and  so  on.  The  last  divisor  thus  found  will  be  the 
G.C.D.  sought. 

7.  Since  the  G.C.D.  of  two  or  more  quantities  contains  all  the  fac- 
tors common  to  these  quantities,  it  may  often  be  found  most  easily 
by  separating  the  quantities  into  factors. 

Find  the  G.C.D.  of  the  following  quantities  : 

5.  ba2-\-bax  and  a2 — x2 Ans.  a-\-x. 

6.  x3 — a2x  and  x3 — a3 Ans.  x — a. 

7.  x3 — c2x  and  x'2-{-2cx-\-c2 Ans.  x-\-c. 

8.  x2-\-2x — 3  and  #2-f-5.c-f6 Ans.  sc-j-3. 

9.  6a2+ lla.r+ 3a;2  and  6a2-{-1ax—  Sx2.  Ans.  2a+3x. 

10.  a* — x4  and  a3-\-a2x — ax1 — x3.      .      .     .  Ans.  a2 — x2. 

11.  a2 — 5ax-\-4x2  and  a3 — a2x-\-  Sax2 — Sx3.  Ans.  a — a*. 

12  a'2x* — ahf  and  xh-\-x3y2 Ans.  x2-\-y2. 

13.  ah — x5  and  a13 — xVi        Ans.  a — x. 


G2  KAY'S  ALGEBRA,  FIRST  BOOK. 


LEAST     COMMON     MULTIPLE. 

107.  A  Multiple  of  a  quantity  is  that  which  contains 
it  exactly. 

Thus,  6  is  a  multiple  of  2,  or  of  3  ;  and  24  is  a  mul- 
tiple of  2,  3,  4,  etc.  ;  also,  8a263  is  a  multiple  of  2a,  of 
2a2,  of  2a26,  etc. 

108.  A  Common  Multiple  of  two  or  more  quantities 
is  one  that  will  exactly  contain  each  quantity. 

Thus,  12  is  a  common  multiple  of  2  and  3  ;  and  6ax  is 
a  common  multiple  of  2,  3,  a,  and  x. 

109.  The  Least  Common  Multiple  of  two  or  more 
quantities  is  the  least  quantity  that  will  contain  them 
exactly. 

Thus,  6  is  the  least  common  multiple  of  2  and  3 ;  and 
lOxj/  is  the  least  common  multiple  of  2x  and  by. 

Note  . — L.C.M.  stands  for  least  common  multiple. 

Remark. — Two  or  more  quantities  can  have  but  one  L.C.M., 
while  they  may  have  an  unlimited  number  of  common  multiples. 

HO.  To  find  the  L.C.M.  of  two  or  more  quantities. 

It  is  evident  that  one  quantity  will  not  contain  another 
exactly,  unless  it  contains  all  of  its  prime  factors. 

Thus,  30  does  not  exactly  contain  14,  because  30—  2X^X5,  and 
14=2x7;  the  prime  factor  7  not  being  one  of  the  prime  factors 
of  30.  It  contains  6,  because  6=2x3,  the  prime  factors  2  and  3  be- 
ing common  to  both  numbers. 

111.  In  order  that  any  quantity  may. exactly  contain 
two  or  more  quantities,  it  must  contain  all  the  different 
prime  factors  of  those  quantities.  And,  to  be  the  least 
quantity  that  shall  exactly  contain  them,  it  should  contain 
these  different  prime  factors  only  once,  and  no  other  factors 
beside. 


ax 

bx 

abc 

X 

bx 

be 

1 

b 

be 

1 

1 

e 

LEAST  COMMON  MULTIPLE.  63 

Thus,  the  L.C.M.  of  a2bc  and  aex,  is  a2bcx,  since  it  contains  all 
the  factors  in  each  of  these  quantities,  and  does  not  contain  any 
otiier  factor. 

With  this  principle,  let  us  find  the  L.C.M.  of  ax,  bx, 
and  abc. 

Arranging  the  quantities  as  in  the  margin,  we 
X    X     bx        be      see  tnat  a  *s  a  factor  common  to  two  of  the  terms; 
hence,  it  must  be  a  factor  of  the  L.C.M.,  and  we 
place  it  on  the  left. 

We  then  cancel  this  factor  in  each  of  the  quan- 
tities in  which  it  is  found,  by  dividing  by  it,  so  that  it  may  occur 
but  once.  As  x  is  a  common  factor  in  the  first  and  second  terms, 
we  divide  by  X  for  the  same  reason,  and  then  by  b. 

We  thus  find,  that  a,  x,  6,  and'c  are  all  the  prime  factors  in 
the  given  quantities;  therefore,  their  product,  abcx,  will  be  the 
L.C.M.  of  these  quantities.     Hence, 


TO    FIND   THE  LEAST    COMMON    MULTIPLE  OF  TWO   OR  MORE 
QUANTITIES, 

Rule. — 1.  Arrange  the  quantities  in  a  horizontal  line. 

2.  Divide  them  by  any  prime  factor  that  will  divide  two 
or  more  of  them  without  a  remainder,  and  set  the  quotients, 
together  with  the  undivided  quantities,  in  a  line  beneath. 

3.  Continue  dividing  as  before,  until  no  prime  factor,  ex- 
cept unity,  will  divide  two  or  more  of  the  quantities  without 
a  remainder. 

4.  Multiply  the  divisors  and  the  quantities  in  the  last  line 

Review. — 107.  What  is  a  multiple  of  a  quantity?  Example. 
108.  A  common  multiple  of  two  or  more  quantities?     Example. 

109.  What  is  the  L.C.M.  of  two  or  more  quantities  ?  Example. 
How  many  common  multiples  may  a  quantity  have? 

110.  When  is  one  quantity  not  contained  exactly  in  another? 
Example.  111.  When  contained  exactly?  Example.  What  is 
necessary,  in  order  that  one  quantity  may  exactly  contain  two  or 
more  quantities? 

111.  What  is  necessary,  in  order  that  any  quantity  may  be  the 
least,  that  shall  contain  two  or  more  quantities  exactly?  What  fac- 
tors does  the  L.C.M.  of  two  or  more  quantities  contain?  Rule  for 
finding  the  L.C.M.  ?     For  finding  it  by  separating  into  factors? 


64  RAY'S  ALGEBRA,  FIRST  BOOK. 

together,  and  the  product  will  be  the  least  common  midtiple 
required. 

Or,  Separate  the  quantities  into  their  prime  factors ;  then, 
to  form  a  product,  1st,  take  each  factor  once;  2d,  if  any 
factor  occurs  more  than  once,  take  it  the  greatest  number  of 
times  it  occurs  in  either  of  the  quantities. 

112.  Since  the  G.C.D.  of  two  quantities  contains  all  the 
factors  common  to  them,  it  follows,  that  if  we  divide  the 
product  of  two  quantities  by  their  G.C.D.,  the  quotient  will 
be  their  L.C.M. 

Find  the  L.C.M.  in  each  of  the  following  examples : 

1.  4a2,   Sa3x,  and  6ao>Y. '    Ans.  12a3xy. 

2.  12a2x2,  6a3,  and  Sx4y2 Ans.  2±o?xiy\ 

3.  15,  6xz2,  9x2z\  and  IScx3 Ans.  90cxV. 

4.  4a*(a— »),  and  6ax\a2— x2).         Ans.  12a2x\a2— x2). 

5.  10a2x2(x—y),  lhx\x-\-y),  and  12(V— y2). 

Ans.  60a2x\x2— y2). 

GENERAL    REVIEW. 

Note. — These  General  Reviews  are  not  intended  to  be  full  and 
exhaustive,  but  simply  suggestive  to  the  teacher,  who  can  extend 
the  questions,  making  them  as  thorough  and  complete  as  is  deemed 
desirable. 

Define  mathematics.  Quantity.  Algebra.  Arithmetic.  When 
is  quantity  called  magnitude?  When  multitude?  Define  a  prob- 
lem. Theorem.  Known  quantities.  Unknown.  How  are  known 
quantities  represented?     Unknown? 

Name  the  principal  signs  used  in  algebra.  Define  factor. 
Coefficient.  Power.  Exponent.  Root.  A  monomial.  Binomial. 
Trinomial.  Polynomial.  Residual  quantity.  Numerical  value. 
Dimension  of  a  term.     Degree.     Reciprocal  of  a  quantity. 

Define  addition.  Algebraic  subtraction.  What  the  difference 
between  algebraic  and  arithmetical  subtraction?  Define  multipli- 
cation. Division.  What  the  rule  of  the  Coefficients?  Rule  of  the 
exponents?  Rule  for  the  signs?  Illustrate  Theorem  I.  Theorem 
II.;  III.;  IV.;  V.;  VI.;   VII.;  VIII. 

What  the  divisoT  of  a  quantity?  A  prime  number?  Compos- 
ite? A  prime  quantity?  Composite?  Quadratic  trinomial?  A 
common  divisor?  Greatest  common  divisor?  Illustrate  Proposi- 
tion I.;  II.;  III.  Rule  for  the  G.C.D.  Define  a  multiple.  Common 
multiple.     Least  common  multiple.     Rule  for  the  L.C.M. 


ALOKKKAIC  FRACTIONS.  65 

III.    ALGEBRAIC    FRACTIONS. 

DEFINITIONS  AND  FUNDAMENTAL   PROPOSITIONS. 

113.  A  Fraction  is  an  expression  representing  one  or 
more  of  the  equal  parts  into  which  a  unit  is  supposed  to 
be  divided. 

Thus,  if  the  line  A  B  be  supposed  c        d        e 

to  represent  1  foot,  and  it  be  divided      -*   J     —  ]  I  ■  ■" 

into  4  equal  parts,  1  of  those  parts,  as  Ac,  is  called  one  fourth  (1); 
2  of  them,  as  Ad,  are  called  two  fourths  (|) ;  and  3  of  them,  as  Ae, 
are  called  three  fourths  (|). 

In  the  algebraic  fraction  -,  if  C=4  and  1  denotes  1  foot,  then  - 

denotes   one  fourth   of    a   foot.     In    the  fraction    -,   if   a=3   and 

11  a  ° 

-=j  of  a  foot,  then  -  represents  three  fourths  (|)  of  a  foot. 

114.  An  Entire  Algebraic  Quantity  is  one  not  ex- 
pressed under  the  form  of  a  fraction. 

Thus,  ax-\-b  is  an  entire  quantity. 

115.  A  Mixed  Quantity  is  one  composed  of  an  entire 
quantity  and  a  fraction. 

b  ,  • 

Thus,  a-] —  is  a  mixed  quantity. 

116.  An  Improper  Algebraic  Fraction  is  one  whose 

numerator  can  be  divided  by  the  denominator,  either  with 

or  without  a  remainder. 

ab            ax2-\-b 
Thus,   —  and  are  improper  fractions. 

117.  A  Simple  Fraction  is  a  single  factional  expres- 
sion  ;    as,   o,  j,  or  -?.    It  may  be  either  proper  or  improper. 

Review. — 112.  If  the  product  of  two  quantities  be  divided  by 
their  G.C.D.,  what  will  the  quotient  be? 

113.  What  is  a   fraction?      114.  An  entire  algebraic  quantity? 
Example.     115.  A  mixed  quantity?     Example.     110.  An  improper 
algebraic  fraction  ?     Example. 
1st  Bk.  6 


66  RAYS  ALGEBRA,  FIRST  BOOK. 

118.  A  Compound  Fraction  is  a  fraction  of  a  fraction; 

1  2  m         a 

as,  n  of    O)  or  ~~  °f    i* 
'2  o'         n  b 

119.  A  Complex  Fraction  is  one  that  has  a  fraction 
either  in  its  numerator  or  denominator,  or  in  both. 

Z-    d-    a-\-~  a-\-~ 

2      2         c  c   • 

Thus,   -4,   -^,  — 77~j  an^    — m->  are  complex  fractions. 

3  ^ 

120.  Algebraic  fractions  are  represented  in  the  same 
manner  as  common  fractions  in  Arithmetic. 

The  Denominator  is  the  quantity  below  the  line,  and  is 
so  called  because  it  denominates  or  shows  the  number  of 
parts  into  which  the  unit  is  divided. 

The  Numerator  is  the  quantity  above  the  line,  and  is 
so  called  because  it  numbers  or  shows  how  many  parts  are 
taken. 

Thus,  in  the  fraction  |,  it  is  understood  that  the  unit  is  divided 
into  4  equal  parts,  and  that  three  of  these  parts  are  taken :  -  de- 
notes that  a  unit  is  divided  into  C  equal  parts,  and  that  a  of  these 
parts  are  taken.      . 

The  numerator  and  denominator  are  called  the  terms  of 
a  fraction. 

121.  In  the  preceding  definitions  of  numerator  and 
denominator,  reference  is  had  to  a  unit  only.  This  is  the 
simplest  method  of  considering  a  fraction ;  but  there  is 
another  point  of  view  in  which  it  is  proper  to  examine  it. 

If  required  to  divide  3  apples  equally,  between  4  boys,  it  can  be 
effected  by  dividing  each  of  the  3  apples  into  4  equal  parts,  and 

Review. — 117.  What  is  a  simple  fraction?  Example.  118.  A  com- 
pound fraction  ?     Example.     119.  A  complex  fraction  ?     Example. 

120.  In  fractions,  what  is  the  quantity  below  the  line  called  ? 
Why?  Above  the  line?  Why?  Example.  What  are  the  terms  of 
a  fraction  ? 


ALGEBRAIC  FRACTIONS.  07 

giving  to  each  boy  3  parts  from  1  apple,  or  1  part  from  each  of 
the  3  apples ;  that  is,  |  of  1  unit  is  the  same  as  i  of  3  units. 

Thus,  |  may  be  regarded  as  expressing  two  fifths  of  one  thing,  or 
one  fifth  of  two  things.  « 

771  1 

So,  —  is  either  the  fraction  -  of  one  unit  taken  m  times,  or  it  is 
'   n  n  ' 

the  nth  of  m  units.  Hence,  the  numerator  may  be  regarded  as 
showing  the  number  of  units  to  be  divided;  and  the  denominator,  as 
showing  the  divisor,  or  what  part  is  taken  from  each. 

122*  Proposition  I. — If  we  multiply  the  numerator  of 
a  fraction  without  changing  the  denominator,  the  value  of  the 
fraction  is  increased  as  many  times  as  there  are  units  in  the 
multiplier. 

If  we  multiply  the  numerator  of  the  fraction  2  by  3,  without 
changing  the  denominator,  it  becomes  ^. 

Now,  2  and  ^  have  the  same  denominator,  which  expresses  parts 
of  the  same  size;  but  the  second  fraction,  £,  having  three  times  as 
many  of  those  parts  as  the  first,  is  three  times  as  large.  The  same 
may  be  shown  of  any  fraction  whatever. 

123.  Proposition  II. — If  we  divide  the  numerator  of  a 
fraction  without  changing  the  denominator,  the  value  of  the 
fraction  is  diminished  as  many  times  as  there  are  units  in  the 
divisor. 

If  we  take  the  fraction  |,  and  divide  the  numerator  by  2,  without 
changing  the  denominator,  it  becomes  |. 

Now,  |  and  3  have  the  same  denominator,  which  expresses  parts 
of  the  same  size ;  but  the  second  fraction,  |,  having  only  one  half 
as  many  of  those  parts  as  the  first,  4,  is  only  one  half  as  large. 
The  same  may  be  shown  of  other  fractions. 

124.  Proposition  III. — If  we  multiply  the  denominator 
of  a  fraction  without  changing  the  numerator,  the  value  of 
the  fraction  is  diminished  as  many  times  as  there  are  units  in 
the  multiplier. 

Review. — 121.  In  what  two  different  points  of  view  may  every 
fraction  be  regarded?  Examples.  122.  How  is  the  value  of  a  frac- 
tion affected  by  multiplying  the  numerator  only  ?     Give  the  proof. 

123.  How  is  the  value  of  a  fraction  affected  by  dividing  the  nu 
merator  only?     Give  the  proof. 


68  RAY'S  ALGEBRA,  FIRST  BOOK. 

If  we  take  the  fraction  |,  and  multiply  the  denominator  by  2,  with- 
out changing  the  numerator,  it  becomes  |. 

Now,  the  fractions  |  and  |  have  the  same  numerator,  which  ex- 
presses the  same  number  of  parts;  but,  in  the  second,  the  parts 
being  only  one  half  the  size  of  those  in  the  first,  the  value  of  the 
second  fraction  is  only  one  half  that  of  the  first.  The  same  may  be 
shown  of  any  fraction  whatever. 

125.  Proposition  IV. — If  we  divide  the  denominator  of 
a  fraction  without  changing  the  numerator,  the  value  of  the 
fraction  is  increased  as  many  times  as  there  are  units  in  the 
divisor. 

If  we  take  the  fraction  3  and  divide  the  denominator  by  3  with- 
out changing  the  numerator,  it  becomes  |. 

Now,  the  fractions  %  and  |  have  the  same  numerator,  which 
expresses  the  same  number  of  parts ;  but,  in  the  second,  the  parts 
being  three  times  the  size  of  those  of  the  first,  the  value  of  the 
second  fraction  is  three  times  that  of  the  first.  The  same  may  be 
shown  of  other  fractions. 

126.  Proposition  V.— Multiplying  both  terms  of  a  frac- 
tion by  the  same  number  or  quantity,  changes  the  form  of  a 
fraction,  but  does  not  alter  its  value. 

If  we  multiply  the  numerator  of  a  fraction  by  any  number,  its 
value,  Prop.  I.,  is  increased,  as  many  times  as  there  are  units  in  the 
multiplier;  and,  if  we  multiply  the  denominator,  the  value,  Prop.  III., 
is  decreased,  as  many  times  as  there  are  units  in  the  multiplier. 
Hence. 

The  increase  is  equal  to  the  decrease,  and  the  value  remains  un- 
changed. 

127.  Proposition  VI. — Dividing  both  terms  of  a  frac- 
tion by  the  same  number  or  quantity,  changes  the  form  of  the 
fraction  but  not  its  value. 

If  we  divide  the  numerator  of  a  fraction  by  any  number,  its 
value,  Prop.  II.,  is  decreased,  as  many  times  as  there  are  units  in  the 

Review. — 124.  How  by  multiplying  only  the  denominator?  How 
proved  ?     125.  By  dividing  the  denominator  only  ?     How  proved  ? 

126.  How  is  a  fraction  affected  by  multiplying  both  terms  by 
the  same  quantity?     Why? 


ALGEBRAIC  FRACTIONS.  69 

divisor;  and,  if  we  divide  the  denominator,  the  value,  Prop.  IV.,  i3 
increased  as  many  times.     Hence, 

The  decrease  is  equal  to  the  increase,  and  the  value  remains  un- 
changed. 

CASE    I. 
TO    REDUCE   A    FRACTION   TO    ITS    LOWEST    TERMS. 

128.  Since  the  value  of  a  fraction  is  not  changed  by- 
dividing  both  terms  by  the  same  quantity,  Art.  127,  we 
have  the  following 

Rule. — Divide  both  terms  by  their  greatest  common  divisor. 
Or,  Resolve  the  numerator  and  denominator  into  their  prime 
factors,  and  then  cancel  those  factors  common  to  both  terms. 

Remark. — The  last  rule  is  generally  most  convenient. 

4a&2         .       , 
1.  Reduce   yrr-o  to  its  lowest  terms. 

4:ab2_2abx2b__2ab 
Reduce  the  following  fractions  to  their  lowest  terms : 


o    4«V  A        2xl 

2.  —^ .  .     .  Ans.  —— . 

6  a4  oa 

3.  £*£    .  .     .  Ans.  f? 
Sax3  4x 

4.  9a^  .     .  Ans.  & 
12x:yz*  4y 


r    12.T2yV  A         Sfz 

5. 2 ,     .     .    Ans.  -4— 

%xh*  2  ' 

n    2a'zcx--\-2acx    »  CKC-f-1 

b. ! .  Ans.  — — ! —  • 

10ac2x  DC 

7.  5a'&+5«^       Ans.  M^ 

babc-\-babd  c-j-d' 


8   12.^-18^  Ang    2x-Sy 

'  18afy+l2a;/     '     '     '  3x+2y' 

Note. — In  the  preceding  examples,  the  greatest  common  divisor 
in  each  is  a  monomial;  in  those  which  follow,  it  is  a  polynomial. 

Review. — 127.  How  by  dividing  both  terms  by  the  same  quan- 
tity?    Why?     128.  How  reduce  a  fraction  to  its  lowest  terms? 


70  RAY'S  ALGEBRA,  FIRST  BOOK. 

n     3a3 — Sab2      mi-     •  1   * 

9. This  is  equal  to 

bab-\-bb2  " 

3a(a2— b2)__Sa(a+b)(a— b)_Sa(a— b)    Ang 
bb(a+b)  ~        bb(a+b)  ~bb 


10.  **-***+   9  ,    An,  f 

4^—82^-1-12  4 

11.  ^2-2^+1.     An,!^I. 

»?— 1  »+l 


12.  ^=^!.         An,  JL 


13.  ^      .  An,  !±t 

x2 — 2xy-\-y2  x — # 

-,  ,i         #3 — ace2         A  cc2 

14. Ans. 

xl — 2ax-\-a2  x — a 


15.  *2+2*-15t  Ans<  x- 


3 

a^+y*["    '  a2-f-8x-r-15 ""      '  x+3' 

120.  Exercises  in  division,  Art.  76,  in  which  the  quo- 
tient is  a  fraction,  and  capable  of  being  reduced  to  lower 
terms. 

bx 

1.  Divide  bx2y  by  oxy2 Ans.  -g-. 

2.  Divide  ara?i2  by  a2m2n Ans.  

am 

So,  also,  when  one  or  both  of  the  quantities  are  polynomials. 

3.  Divide  3*f»*-f-S*'  by  15ra2+15n2 Ans.  \, 

xv 

4.  Divide  xzy2-\-xhf  by  ax2y-\-axy2 Ans.  — . 

a 

x—1 

5.  Divide  x2-\-2x — 3  by  x2-\-bx-\-Q.  .     .     .    Ans.         ~. 

x~\~  A 

CASE    II. 


TO    REDUCE    A    FRACTION    TO    AN    ENTIRE    OR     MIXED 
QUANTITY. 

ISO.  Since  the  numerator  of  the  fraction  may  be  re- 
garded as  a  dividend,  and  the  denominator  as  a  divisor, 
this  is  merely  a  case  of  division.     Hence, 


REDUCTION  OF  FRACTIONS.  71 

Rule. — Divide  the  numerator  by  the  denominator  for  the 
entire  part;  and,  if  there  be  a  remainder,  place  it  over  the 
denominator  for  the  fractional  part. 

Note. — The  fractional  part  should  be  reduced  to  its  lowest  terms. 
Reduce  the  following  to  entire  or  mixed  quantities : 

_,     3ax+b2                                         .         Sax-\-b2      0     ,  b2 
1. Ans. — =6a-\ — 

2.   — Ans.  b-\ — 

a  a 

0    a2-\-x2                                                       .             .       .     2x2 
6. . Ans.  a-\-x-\- 


a — x 


A     2a2x — x3  .         n  x3 

4.  Ans.  lax 

a  a 

t     4ax—2x2—a2  .         a  "2 

5.  - — -~ Ans.  Zx- 


2a — x  2a — i 

O.   j Ans.  a2 — ax-\-xl 


H        12^— Sx2  .        o  ,      3 

7.  T-= T—A — r-r Ans.  3+-— 


a-\-x  a-\-x 

\2x?—3x2 
4x3 — x2 — 4x+l* 


CASE    III. 

TO   REDUCE   A   MIXED   QUANTITY   TO   THE    FORM    OF   A 
FRACTION. 

131. — 1.  In  2\,  there  are  how  many  thirds? 

In  1  unit  there  are  3  thirds;  hence,  in  2  units  there  are  6  thirds; 
then,  21  or  2+J=J  +  J=J. 

_  b    ac     b    acA-b        ,        b    ac     b    ac — b 

bo,  a  A — == 1 — = ;  and  a = = ,     Hence, 

c      c      c         c  c      c      c         c 


TO    REDUCE    A    MIXED    QUANTITY    TO   THE    FORM    OF    A 
FRACTION, 

Rule. — Multiply  the  entire  part  by  the  denominator  of  the 
fraction.     Add  the  numerator  of  the  fractional  part  to  this 


72  RAYS  ALGEBRA,  FIRST  BOOK. 

product,  or  subtract  it,  as  the  sign  may  direct,  and  place  the 
result  over  the  denominator. 

Remark. — Cases  II.  and  III.  are  the  reverse  of,  and  mutually 
pi-ove  each  other. 

Before  proceeding  further,  it  is  important  to  consider 


THE     SIGNS     OF     FRACTIONS. 

1«£2.  The  signs  prefixed  to  the  terms  of  a  fraction 
affect  only  those  terms ;  but  the  sign  placed  before  the 
dividing  line  of  a  fraction,  affects  its  whole  value. 

a2—b2 

Thus,  in ,  the  sign  of  a2,  in  the  numerator,   is  plus; 

'  x^-y  '  fe  '  '        F      ' 

of  62,  minus;  while  the  sign  of  each  term  of  the  denominator  is 

plus.     But  the  sign  of  the  fraction,  taken  as  a  whole,  is  minus. 

-i-ab 
By  the  rule  for  the  signs  in  division,  Art.  75,  we  have =-}-&; 

or,  changing  the  signs  of  both  terms, =-\-b. 

—Ct 

Changing  the  sign  of  the  numerator,  we  have •=== — b. 

-+-01/ 

Changing  the  sign  of  the  denominator,  we  have =» — b. '   Hence, 

— ct  » 

The  signs  of  both  terms  of  a  fraction  may  be  changed 
without  altering  its  value  or  changing  its  sign;  but,  if  the  sign 
of  either  term  of  a  fraction  be  changed,  and  not  that  of  the 
other,  the  sign  of  the  fraction  will  be  changed. 

Hence,  The  signs  of  either  term  of  a  fraction  may  be 
changed,  without  altering  its  value,  if  the  sign  of  the  fraction 
be  changed  at  the  same  time. 

Review. — 130.  How  reduce  a  fraction  to  an  entire  or  mixed 
quantity?     131.  A  mixed  quantity  to  the  form  of  a  fraction? 

132.  What  do  the  signs  prefixed  to  the  terms  of  a  fraction  affect? 
The  sign  placed  before  the  whole  fraction?  What  is  the  effect  of 
changing  the  signs  of  both  terms  of  a  fraction?  Of  one  term,  and 
not  the  other?     The  sign  of  the  fraction,  and  of  one  of  its  terms? 


REDUCTION  OF  FRACTIONS.  73 

ax — x2        ax — x2       x2 — ax 


Thus,    .     . 

c  — c  c 

.    .  a—x        ,  a— x        ,  x—a 

And,    .    .    .    a-  -T-=a+-—^=a+—5-. 


fin/' rr 

1.  Reduce   3a-f- to  a  fractional  form. 

x 

Sax       .  Sax  ,  ax — a    Sax-\-ax — a    4ax — a 

Sa= and as ■ = .     Ans. 

x  x     '      x  x  x 

2.  Reduce  4a ~ —  to  a  fractional  form. 

6c 

12ac       ,  12ac     a—b    12ac—(a—b)    I2ac—a+b 
4o=-TkT'  wi  -55 33"= S3 = 35 ■  Ans- 


Reduce  the  following  quantities  to  improper  fractions  : 

3.  2-ff  and  2— § Ans.  ^  and  }, 

4.  5c+t^ Ans.  10™+"-\ 

JLx  Ax 

-     -        a — b  .         10cx — a-\-b 

b'bc 2x~ AnS' Tx ' 

a    0      4x2— 5  .       llx2+5 

b.  ox = Ans.  — = . 

ox  ox 

K    0    .  3a — v2  39y2-f-3a 

r8»+-w~ Ans-^^- 

8.  ,_i+J=! Ads'JtZ 

9.  ^     5 Ans.  4.y-10x-5, 

10.  — ^ 6 Ans.  ~ — . 

-•-  .  a2±x2— 5  2  2a2— 5 

11.  a — x-\- j Ans.   -. — . 

a-\-x  •  a-f-x 

12.  a3— a*x+ax*— x*—  °-^- Ans. ^~ 

a+x  a+a> 

1st  Bk.  7* 


RAYS  ALGEBRA,  FIRST  BOOK. 


CASE    IV. 

TO    REDUCE    FRACTIONS    OF    DIFFERENT    DENOMINATORS    TO 
EQUIVALENT   FRACTIONS    HAVING   A   COMMON 
DENOMINATOR. 
Oj  c 

133. — 1.  Reduce  -  and  -,  to  a  common  denominator. 
b  a 

Multiply  both  terms  of  the  first  fraction  by  d,  the  denominator 
of  the  second,  and  both  terms   of  the  second  fraction  by  b,  the 

denominator  of  the  first.    We  shall  then  have  -p-j  and  j-=. 

bd         bd 

In  this  solution,  observe ;  first,  the  values  of  the  fractions  are  not 
changed,  since,  in  each,  both  terms  are  multiplied  by  the  same 
quantity;   and, 

Second,  the  denominators  must  be  the  same,  since  they  consist  of 
the  product  of  the  same  quantities. 

2.  Reduce  — ,  -,  and  -,  to  a  common  denominator, 
m    n  r 

Here,  we  multiply  both  terms  of  each  fraction  by  the  denominators 
of  the  other  two  fractions.         Thus, 

aynyr      anr      by^my^r     bmr      cymyn     cmn 
"mXnXr      mnr'    ny(rnyr~ mnr '    ry^myn     mnr' 

It  is  evident  that  the  value  of  each  fraction  is  not  changed,  and 
that  they  have  the  same  denominators.     Hence, 


TO    REDUCE    FRACTIONS   TO   A   COMMON   DENOMINATOR, 

Kule. — Multiply  both  terms  of  each  fraction  by  the  prod- 
uct of  all  the  denominators  except  its  own. 

Remark  . — This  is  the  same  as  to  multiply  each  numerator  by 
all  the  denominators  except  its  own,  for  the  new  numerators;  and  all 
the  denominators  together,  for  the  common  denominator. 

Review. — 133.  How  do  you  reduce  fractions  of  different  denomi- 
nators to  equivalent  fractions  having  the  same  denominator? 

Why  is  the  value  of  each  fraction  not  changed  by  this  process? 
Why  does  this  process  give  to  each  fraction  the  same  denominator? 


REDUCTION  OF  FRACTIONS..  75 

Reduce  to  fractions  having  a  common  denominator : 

_    a     c          _  1  .  2ad     26c  bd 

B'  V  d'  and  2 Aas'  VM  W®  and  2bd 


4. 


x         .  x-\-a  K        ex         _  xy+ay 

-,  and  — — Ans.  — ,  and    J       J. 


y  c  cy  cy 


.  2     3a         .  a— -y  .          86      9«6          ,    12a:— 12y 

5.  gj   T,  and  -J.  Ans.  m    j-,  and  — ^ 

^  2.x     3#         ,  .          10.T2      9xy           '    lbayz 

O.  tt-,    -=—y  and  a.     .  Ans.  =-^ — ,    .,    *  ,  and 


3y'   5*'                                '  15^'    Ibyz1  \byz 

7.  — — ,  and  — -^.    Ans.  —    ,       '      ,  and ,       . 


a— y  #+#  x2— yz  xl—y- 

o         36     ,         .   -  .         ac    36    cd 

8.  a,  — ,  d,  and  5.     .     .     .  Ans.  — ,    — ,   — ,  and 


134.  When  the  denominators  of  the  fractions  to  be 
reduced  contain  one  or  more  common  factors,  the  preced- 
ing rule  does  not  give  the  least  common  denominator. 

If  we  find  the  L.C.M.  of  all  the  denominators,  and  divide 
it  by  the  denominators  severally,  it  is  easy  to  see  that  we 
shall  obtain  multipliers  for  each  of  the  fractions,  which 
will,  without  changing  their  value,  make  their  denomina- 
tors the  same  as  the  L.C.M. 


1.  Reduce  -r.  T-,    and  — 7,  to  equivalent  fractions  hav- 
b    bo,  cd  ^ 

ing  the  least  common  denominator. 

The  L.C.M.  of  the  denominators  is  bod;  dividing  this  by  b,  bc} 

and  cd,  we  obtain  cd,  d,  and  b.     Multiplying  both  terms  by  these 

incd    nd  br 

severally,  we  have  -^,  ^  and  ^;  or  thus: 

a  ^    i,      ^       j  m     cd  med 

ocd-i-o=cd,  and  -=-V — r57*    ■     •    •       *— -3* 

b  /  cd  bed 

bed—be— d,  and  ■*-     -,=  ....       -= — » 
6cxd  bed 

bed-^-cd—b,  and  — /\=  •  •       *-t5 

cdx6  6cd 


76 

The  process  of  multiplying  the  denominators  may  be  omitted,  as 
the  product  in  each  case  is  the  same.     Hence, 

TO    REDUCE    FRACTIONS.  OF    DIFFERENT    DENOMINATORS    TO 
EQUIVALENT  FRACTIONS   HAVING  THE   LEAST 
COMMON    DENOMINATOR, 

Rule. — 1.  Find  the  least  common  multiple  of  all  the  de- 
nominators; this  will  be  the  common  denominator. 

2.  Divide  the  least  common  multiple  by  the  first  of  the 
given  denominators,  and  multiply  the  quotient  by  the  first  of 
the  given  numerators ;  the  product  will  be  the  first  of  the 
required  numerators. 

3.  Proceed,  in  a  similar  manner,  to  find  each  of  the  other 
numerators. 

Note  . — Each  fraction  should  first  be  reduced  to  its  lowest  terms. 

Reduce  the  following  to  equivalent  fractions  having  the 
least  common  denominator : 

0     2a     Sx  Sy  .  4ad     lSbx        ,    bey 

2-  36?   Vd'  and  Wd-    ■    Ans-  SMC  66^'  and  66^- 

0    m       n        r  b2cdm      acdn        ab2r 

o.   — ,    n-,    -T-=.     .     .     .       Ans 


a&W    aVM    ab2c2d' 
4    x+y    x—y    x2+y2     ^    (x-\-y)2     (x—y)2    x2-{-y2 
x — y*    a;+y    xl — y2'  '    x2 — y2  '     x1 — y2  '    x2 — y2' 

Note. — The  two  following  Art's  will  be  of  frequent  use,  par- 
ticularly in  completing  the  square,  in  the  solution  of  equations  of 
the  second  degree. 

135.  To  reduce  an  entire  quantity  to  the  form  of  a 
fraction  having  a  given  denominator. 

1.  Let  it  be  required  to  reduce  a  to  a  fraction  having 
b  for  its  denominator. 

Since  a=-=,  if  we  multiply  both  terms  by  b,  which  will  not 
change  its  value,  Art.  126,  we  have  -^=-=-.     Hence, 


REDUCTION  OF  FRACTIONS.  77 

TO     REDUCE     AN     ENTIRE    QUANTITY    TO     THE    FORM    OF    A 
FRACTION    HAVING   A   GIVEN    DENOMINATOR, 

Rule> — Multiply  the  entire  quantity  by  the  given  denomina- 
tor, and  write  the  product  over  it. 

2.  Reduce  x  to  a  fraction  whose  denominator  is  4.  Ans.  -j  . 

3.  Reduce  m  to  a  fraction  whose  denominator  is  9a2. 

9a2m 
Ans'  1ST: 

4.  Reduce  3c-f-5  to  a  fraction  whose  denominator  is  16c2. 

48c3+80c2 

Ans. jTn . 

16c2 

5.  Reduce    a — b   to    a    fraction    whose    denominator    is 

a3_3a26_|_  3afc2__£3     (a_iy 


a2—2ab-\-b\  Ans. 


a2— 2ab+b2  (a— by' 


13©.  To  convert  a  fraction  to  an  equivalent  one,  hav- 
ing a  denominator  equal  to  some  multiple  of  the  denom- 
inator of  the  given  fraction. 

1.  Reduce  T  to  a  fraction  whose  denominator  is  be. 
b 

It  is  evident  that  this  will  be  accomplished  without  changing  the 
value  of  the  fraction,  by  multiplying  both  terms  by  C.  This  multi- 
plier, c,  may  be  found  by  inspection,  or  by  dividing  be  by  b. 
Hence, 


TO   CONVERT   A   FRACTION  TO   AN  EQUIVALENT  ONE   HAVING 
A  GIVEN  DENOMINATOR, 

Rule. — Divide  the  given  denominator  by  the  denominator 
of  the  fraction,  and  multiply  both  terms  by  the  quotient. 

Review. — 134.  How  reduce  fractions  of  different  denominators  to 
equivalent  fractions  having  the  least  common  denominator? 

134.  If  each  fraction  is  not  in  its  lowest  terms  before  commencing 
the  operation,  what  is  to  be  done?  135.  How  reduce  an  entire  quan- 
tity to  the  form  of  a  fraction  having  a  given  denominator? 


78  RAY'S  ALGEBRA,  FIRST  BOOK. 

Remark. — If  the  required  denominator  is  not  a  multiple  of  the 
given  one,  the  result  will  be  a  complex  fraction.  Thus,  if  it  is 
required  to  convert  £  into  an  equivalent  fraction  whose  denomina- 
tor is  5,  the  numerator  of  the  new  fraction  would  he  2\. 


2.  Convert  -y  to  an  equivalent  fraction,  having  the  de- 

12 
nominator  16.  Ans.  q-^ 

16* 

3.  Convert  —  to  an  equivalent  fraction,  having  the  de- 
nominator  *£  Ans.  ^ 

171  ~ T~  71/ 

4.  Convert  — to  an  equivalent  fraction,  having  the 

m~n                                                    m2— £ 
denominator  m2 — 2mn-\-n2.  Ans.  — - — ~ ; — - 


CASE     V. 
ADDITION    AND    SUBTRACTION    OF    FRACTIONS. 

137. —  1.  Required  to  find  the  sum  of  §  and  I. 

Here,  both  parts  being  of  the  same  kind,  that  is,  fifths,  we  may 
add  them  together,  and  the  sum  is  6  fifths,  (|). 

2.  Let  it  be  required  to  find  the  sum  of  —  and  — . 

m  m 

Here,   the  parts  being   the  same,   that  is,  withs,   we  shall  have 
a      b     a-\-b 

mm      m .  * 


(t  c 

3.  Let  it  be  required  to  find  the  sum  of  —  and  -. 

in  n 

Here,  the  denominators  being  different,  we  can  not  add  the  nu- 
merators, and  call  them  by  the  same  name.  We  may,  however, 
reduce  them  to  a  common  denominator,  and  then  add. 

Thus,  2UHK    °=*™;  and  ^+^!=fl»+cw.     Hence, 
ra    ma    a    mn  ran    mn        ma 


ADDITION  AND  SUBTRACTION  OF  FRACTIONS.         79 
TO  ADD    FRACTIONS, 

Rule. — Reduce  the  fractions,  if  necessary ,  to  a  common 
denominator;  add  the  numerators  together,  and  write  their 
sum  over  the  common  denominator. 

138.  It  is  obvious  that  the  same  principles  would 
Bpply  in  finding  the  difference  between  two  fractions. 
Hence, 

TO    SUBTRACT    FRACTIONS, 

Rule. — Rediece  the  fractious,  if  necessary,  to  a  common 
denominator;  subtract  the  numerator  of  the  subtrahend  from 
the  numerator  of  the  minuend,  and  write  the  remainder  over 
the  common  denominator. 

EXAMPLES    IN    ADDITION    OF    FRACTIONS. 

4.  Add  ^,  7t,  and  -^  together Ans.  a. 

5.  Add  7j,  ■£,  and  -n  together Ans.  T7r. 

o    o  o  10 

c      ,,,11  ,    1    ,        ,,  A  bc+ac-\-ab 

b.  Add  -,  T,  and  —  together.     .     .  Ans.  = . 

a    b  c  abc 

*     a  1 1   x    V         i    z  x       A!  a         6x-+-4y-\-3z 

7.  Add  3,  ^,  and  ^  together.     .     .  Ans. =-^ • 

8.  Add  -7-,  -f-i  and  -n  together.  Ans.  -7tk-=zscH— 7?*  . 

4      0  o      °  bO  60 

9.  Add  -^J^  and  S?  together Ans.  x. 

A  A 

10.  Add  — r-v  and together.     .     .      .  Ans. 


a-\-b  a—b      &  aL—bl 

11.  Add  -,  -,  and  •«-  together. 

V           ay    ■            3a                         15a+6y+9 
Ans.  q • 

12.  Add  ^~.  _~"I^   and  ^— ^  together.  .     .      Ans.  0. 

ab        be  ac 


80  RAY'S  ALGEBRA,  FIRST  BOOK. 

Entire  quantities  and  fractions  may  be  added  separately ;  or,  the 
entire  quantities  may  be  put  into  the  form  of  fractions  by  making 
their  denominators  unity.  When  mixed  quantities  occur,  it  is  often 
better  to  reduce  them  to  the  form  of  improper  fractions. 

13.  Add  2x,  3^4-^-j  and  ic+^- together.  Ans.  6x-fxF- . 

a  ■,  ■,  r     .  * — 2       ,   .        2x — 3  ,        . 

14.  Add  bx-\-— a—  and  4x — — = together. 

a        a     ,  5a?— 16a+  9 

Ans.  \)x-\-         -,  g . 

loo; 


EXAMPLES    IN    SUBTRACTION    OF    FRACTIONS. 

.,     tp,  a      .     a  a 

1.  ±rom  H  take  77 Ans.  - 

2  6       ■  b 

a    _  3x  x  .      2a;  a; 

Z.  ±  rom  -j-  take  -^- Ans.  ^^f. 

4  3  12 

3.  From  — ~—  take      . Ans.  b. 

A  "A 

.     -,  2ax      .      5ax  .  llax 

4.  Irom  — s-  take  -g- Ans. — -. 

o  A  0 

_    ^  3    ,   .       5  .         Sx — 10a 

5.  From  -r-  take  ?r- Ans.  - — -. . 

4a  Ax  4:ax 

6.  From  *±l  take  £1.    ......       Ans.  -^L. 

a; — y  x-\-y  xl — yl 

H    Lj  2o+5      .      3a— 5  A        126— a 

7.  Irom   — = take — u — -.     .  .     .     Ans. — ^-= — . 

be  7c  35c 

o    -n  -i^.i      n        -T — ^    4        o     .  bx-\-cx — b2 

o.  Irom  ox-\-T  take  Za; .  Ans.  6x-\- - . 

be  be 

11  25 

9.  From  r  take  — — r Ans.  — — =-:. 

a — 6  a-f-6  a1 — bl 

-in    i7  ,  ,    ,   ,      1  1  1  *         tfb+ab* — a—b 

10.  Irom  a-\-b  take  —+-5-.   .     .      Ans. -. . 

a     b  c.b 

Review. — 136.  How  convert  a  fraction  to  an  equivalent  one  hav- 
ing a  gi'-.n  denominator?     Explain  the  operation  by  an  example. 

137.  When  fractions  have  the  same  denominator,  how  add  them 
together?     When  fractions  have  different  denominators? 

138.  If  two  fractions  have  the  same  denominator,  how  find  their 
difference?     When  they  have  different  denominators? 


MULTIPLICATION  OF  FRACTIONS.  gj 

11.  From —  take  — -^-      .     .     .  Ans.  ■ — — £_ 

x—y  x+y  x2—y2     ' 

12.  From  x-\ =-  take  — r-^-.  .     .     .    Ans. 

x — 1  a;  4-1  x 

13.  From  2a—Sx-\-^—^-  take  a — ^x-^—., 

a  x 

Ans.  a-\-2x-\- 


ax 

CASE    VI. 
MULTIPLICATION   OF    FRACTIONS. 

139.  To  multiply  a  fraction  by  an  entire  quantity,  or 
an  entire  quantity  by  a  fraction. 

It  is  evident,  from  Prop.  I.,  Art.  122,  that  this  may  be 
done  by  multiplying  the  numerator. 

a     _    2a  a  am 

Thus,   5X2=^-,  and -gX^^-fc-. 

Again,  as  either  quantity  may  be  made  the  multiplier,  Art.  67,  to 
multiply  4  by  |,  is  the  same  as  to  multiply  |  by  4.     Hence, 

TO  MULTIPLY   A  FRACTION  BY  AN  ENTIRE  QUANTITY,    OR  AN 
ENTIRE   QUANTITY   BY  A    FRACTION, 

Rule. — Multiply  the  numerator  by  the  entire  quantity,  and 
write  the  product  over  the  denominator. 

From  Art.  125,  it  is  evident  that  a  fraction  may  also 
be  multiplied  by  dividing  its  denominator  by  the  entire 
quantity. 

Thus,  in  multiplying  |  by  2,  we  may  divide  the  denominator 
by  2,  and  the  result  will  be  |,  which  is  the  same  as  to  multiply 
the  numerator  by  2,  and  reduce  the  resulting  fraction  to  its  lowest 
terms.    Hence, 

Review. — 139.  How  multiply  a  fraction  by  an  entire  quantity,  or 
an  entire  quantity  by  a  fraction?  When  the  denominator  is  divis- 
ible by  the  entire  quantity,  what  is  the  shortest  method  ? 


82 


RAYS  ALGEBRA,  FIRST  BOOK. 


In  multiplying  a  fraction  and  an  entire  quantity  together, 
we  should  always  divide  the  denominator  of  the  fraction  by 
the  entire  quantity,  when  it  can  be  done  without  a  remainder. 


Remark. — The  expression  u|  of  6"  is  the  same  as  fX^. 


1.  Multiply  —  by  ad. 


2.  Multiply  —^—  by  xy. 

3.  Multiply  a— 2b  by 


2a+c*  ' 

4.  Multiply  a2— b2  by     C~° '.  Ai 

_     ,r   ,,.   ,      2a-\-Sxz  .         ,• 

5.  Multiply  ~ —  by  ab.      . 


Ans. 


2a2d 


be  ' 
Ans.^±^ 


An; 


4ac—Sh 


2a+c    ' 
Sa2c—Sb2c—a^-\-ab2 


Ans. 


2a+Sxz 


a2b 
p    at  !*•   i  bbc+SLft       ,      0  ,  .        .         5bc+3bx 

G-  MuUiply  io*y-i4.*y  ^  *«'   Ans'  K=7iy 

7.   Multiply  Jg    °a!+6jr    ,     by  2(a-h).  Ans.  £*&- 

&t<i— org— «U  5c+4<Z 

Ans. 


9.  Multiply  -  by  c.     . 


c— d  ' 

ac  a 

Ans.  — =a,  or  q 
c  1" 


Remark. — If  a  fraction  is  multiplied  by  a  quantity  equal  to  its 
denominator,  the  product  will  equal  the  numerator. 


10.  Multiply  — j-3  by  c-{-d.  .     . 

m2 — n 

11.  Multiply  ^qr^  by  2z+5y. 


.  Ans.  a — b. 
Ans.  in2 — n2. 


140.  To  multiply  a  fraction  by  a  fraction. 

1.  Let  it  be  required  to  multiply  |-  by  3. 

Since  |  is  the  same  as  2  multiplied  by  i,  it  is  required  to  mul- 
tiply |  by  2,  and  take  i  of  the  product. 


MULTIPLICATION  OF  FRACTIONS.  83 

Now,  4X2=§»  and  |  of  |  is  T8^.     Hence,  the  product  of  |  and  | 
is  A- 

la 

So,  to  multiply  —  by  — ,  multiply  —  by  m,  and  take  —  of  theprod- 

a           ma       ,  1     .  ma    ma     TT 
uct.     -Y«=  — ,  and   -of = .     Hence, 

TO    MULTIPLY    A    FRACTION    BY    A    FRACTION, 

Rule. — Multiply  the  numerators  together  for  a  new  numer- 
ator, and  the  denominators  together  for  a  new  denominator. 

Remarks.— 1.  The  expression   "one  third  of  one  fourth''  has 
the  same  meaning  as  "  I  multiplied  by  l ;?'  or,  l  of  5—^X3- 

2.  If  either  of  the  factors  is  a  mixed  quantity,  reduce  it  to  an 
improper  fraction  before  commencing  the  operation. 

3.  When  the  numerators  and  denominators  have  common  factors, 
indicate  the  multiplication,  and  cancel  such  factors- 

Thus    14^v   5c  -     2XfXg«c     __2ac 
'  Tbbx~21d~  ^X3X3X^<T"  9bd' 

Also,      5a    xa+5—      5tt(a+6)       —      5 


o«_6*^    9a       2a(a-\-b)(a—b)     2(a—b)' 

^     _,  ,  .   .     3a  ,      hx  .         15ax 

1.   Multiply  -£-  by  -g- Ans.  -g^-r 

rt    ,T  .  .   .     4a  ,      3sc  12 

z.  Multiply  -=—  by  ^— Ans.  «-. 

3.  Multiply  3^jl)  by  ~ Ans.  6x. 

4.  Multiply  — ^—  by  -B- Ans.  j- — . 

5.  Multiply  t=f  by      * Ans.  ^=^. 

¥  J       ab        J    x-\-y  b 

<3.  Multiply  -^-  by  ^±^ Ans.  1. 

Review. — 140.  How  do  you  multiply  one  fraction  by  another? 
Explain  by  analyzing  an  example.  When  one  factor  is  a  mixed 
quantity,  what  ought  to  be  done?  When  the  numerator  and  denom- 
inator have  common  factors?  What  the  meaning  of  "one  third  of 
one  fourth?" 


84  RAY'S  ALGEBRA,  FIRST  BOOK. 


7.  Multiply  ~  by  — ^-.  .  Ads. 


xl       "   a1 — x1  x2(a-\-x)' 

IT-  a        nr  ci  a 

8.  Multiply  — - — j — ,  and together.    Ans.  - 

r  *   a-\-x        xl  a — x  x 

9.  Multiply  — ^— ,    — — jjj  an(i  a-\-b  together.  Ans.  1. 

10.  Multiply  ^±^  by  ^.     .     .      Ans.  —  f+?x    , 

*  J  x*—y*         x+y  x2+2xy-\-yl 

11.  Multiply  c+—  by  S=£     .     .     .  Ans.  **£k*>. 

r  J         '    C X      J     05-f-l  354-J 


CASE    VII. 
DIVISION    OF   FRACTIONS. 

141.  To  divide  a  fraction  by  an  entire  quantity. 

It  has  been  shown,  in  Arts.  123  and  124,  that  a  frac- 
tion is  divided  by  an  entire  quantity,  by  dividing  its 
numerator,  or  multiplying  its  denominator. 

Thus,  j  divided  by  2,  or  I  of  J,  is  §;  or,  |-  2—^=^. 

So,  divided  by  m,  or  —  of    —  is  — .     Eeroe, 

TO    DIVIDE    A    FRACTION    BY    AN    ENTIRE    QUANTITY, 

Rule. — Divide  the  numerator  by  the  divisor,  if  it  can  be 
done  without  a  remainder;  if  not,  multiply  the  denominator. 


To  divide  a  number  by  2  is  to  take  ^  of  it,  or  to  multiply  it  by  ^ ; 

to  divide  by  m  is  to  take  —  of  it,  or  to  multiply  it  by  — .     Hence, 
J  m  m 

To  divide  a  fraction  by  an  entire  quantity,  we  may  write 

the  divisor  in  the  form  of  a  fraction,  as  m=r— ,  then  invert 

it,  and  proceed  as  in  multiplication  of  fractions. 

Review. — 141.   How  divide   a   fraction  by  an   entire  quantity? 
Explain  the  reason  of  the  rule  by  analyzing  an  example. 


DIVISION  OF  FRACTIONS. 


85 


1.  Divide  — ^ —  by  Sab. 


2.  Divide 


In 
14ac3m 
llxy 


by  ^lacm} 


3-  Dlvlde  3+2*  by  a- 

4.  Divide  C  ~JTC(-'  by  c-\-d. 


5.  Divide 


c+d 


by  x-\-y. 


6.  Divide  ^  by  6 

3c     J 

7.  Divide  5±4a  by  0-L6. .     . 

a— b      J 

8.  Divide  la+lC  by  2a— 3y. 

h c 

9.  Divide  -— — ,   ,  ,„  by  a — b. 


10.  Divide 


x^—xy+y 


-..  by  cc+y. 


11.  Divide  — Jl — I  by  aA-bc.    . 

6+c       J       ' 

12.  Divide  -^— ; —   by  am — an.  . 


&+, 


13.  Divide 


by  a2+ab+V 


2a 
Ans.  t— 

In 


Ans 
Ans. 


2c2 


llscy" 
g-f-6 

c 
5* 

x+y 


.     .    Ans. 
.    Ans. 
Ans 
Ans. 
Ans 


c+d' 

2a 

36c' 

8-j-5a 

a2—  bT 

3a-f-5c 


Ans. 
Ans. 
.  Ans. 
Ans. 


4aa— 9y 
6— c 


a*—bz' 

x—y 
x*+y*' 

a 

6-j-c. 
ra-j-n 


.  Ans. 


a— b 


1  12.  To  divide  an  integral  or  fractional  quantity  by 
a  fraction. 

1.  How  many  times  is  §  contained  in  4,  or  what  is  .the 
quotient  of  4  divided  by  |  ? 

i  is  contained  in  4  three  times  as  often  as  1  is  contained  in  4,  be- 
cause 1  is  3  times  as  great  as  l;  therefore,  1  is  contained  in  4,  12 
times;  |  is  contained  in  4  only  one  half  as  often  as  l,  since  it  is 
twice  as  great;  therefore,  |  is  contained  in  4,  6  times. 


86  RAY'S  ALGEBRA,  FIRST  BOOK. 


2.  How  many  times  is  —  contained  in  a? 

*  n 

1  .  m 

Reasoning  as  above,  —  is  contained  in  a,  na  times,  and  —  is  con- 

na    .  n  n 

tamed  —  times. 
m 

3.  How  many  times  is  §  contained  in  |  ? 

A  is  contained  in  |,  three  times  as  often  as  1  is  contained  in  I, 
that  is,  |  times ;  and  |,  half  as  often  as  I,  that  is,  |  times. 

.     TT  .  .     m  .,  .      a  _ 

4.  How  many  times  is  —  contained  in  —  r 

J  n  c 

t,         .  ,;   1  .  .,.«cm.  m , 

Reasoning  as  before,  —  is  contained  in  — .  —  times;  and  —  is  con- 

.        an    .  n  c     c  n 

tained  —   times. 
cm 

An  examination  of  each  of  these  examples  will  show  that  the 

process  consists  in  multiplying  the  dividend  by  the  denominator  of 

the  divisor,  and  dividing  it  by  the  numerator.     If,  then,  the  divisor 

be  inverted,  the  operation  will  be  the  same  as  that  in  multiplication 

of  fractions.     Hence, 


TO    DIVIDE    AN    INTEGRAL   OR  FRACTIONAL  QUANTITY    BY  A 
FRACTION, 

Rule. — Invert  the  divisor,  and  proceed  as  in  multiplica- 
tion of  fractions. 

Note. — After  inverting  the  divisor,  abbreviate  by  canceling. 

1.  Divide  4  by  „ Ans.  — 

J    3  a  ' 

2.  Divide  a  by  ^ Ans.  4a. 

«3.  Divide  al?  by  ^— Ans.  ~^- 

J     be  2  • 

a    t^   -j     a  i.     c  a         ^a 

4.  Divide   o  Dy  o ^ns.  q~- 

o         2  oc 

Review. — 142.  How  divide  an  integral  or  fractional  quantity  by 
a  fraction?  Explain  the  reason  of  this  rule,  by  analyzing  the  ex- 
amples given.     When  and  how  can  the  work  be  abbreviated? 


DIVISION  OF  FRACTIONS.  87 


-         ' .,     x2y  .      xy2  .         2hx 

5.  Divide  -g^  by  -&- ■Ans-  tj— . 

da     *    Ab  day 

6.  Divide  — F—   by  =-?■       Ans.  12a. 

5  15 

7.  Divide       T     by  — ^— Ans.  -/. 

5  4?/  5 

0  a2— 7>2         a+6  .        a(a— &) 

8.  Divide  — = —  by  — ■ — Ans.  -*-= — *, 

5  J      a  5 

n    _ .  . .     z2— 4  .     z— 2  ^H-2 

9.  Divide  — j^-  by  — **- Ans.  — g-. 

6  J      2  o 

-i  a    ^-  • !     x2—2xy-\-y2  j,     a— ,y  .        ca — cy 

10.  Divide  f  '  <y    by  -=-£.       .     .       Ans. -. 

1  "I         Tk-      '  1  a  V         «  +  l  A  a 

11.  Divide    - — T  by  — *-^.       .     .     .     Ans.  0 — ■-* 

a2 — 1     J  a— 1  a2-j-2a-f-l 

10    '*.  .,    2z+3-,     lO.c+15  ,       »— y 

12.  Divide  — ^—  by  — .,  '    „ Ans.  — ~ 

x-\-y  xl — yl  0 


i  o     TV   -a      3(a2— *2)         2(a+fc)    .        3(a2— 2ax-\-x2) 

13.  Divide  — by  ~ — — -.Ans.— ~ 

x  a — x  Ax 

-  . '      .  .,        2x2    .         a-  .  2a; 

14.  Divide  -— — .  by  — = — .       .     .     .     Ans.  -= : — 


143.  To  reduce  a  complex  to  a  simple  fraction. 

This  may  be  regarded  as  a  case  of  division,  in  which 

the  dividend  and  the  divisor  are  either  fractions  or  mixed 

quantities. 

2i 
Thus,   st-  is  the  same  as  2}  divided  by  3|. 

b 

*c  b  n 

aw     is  the  same  as  a4—  divided  by  ra-j — . 

.aiso,  n  i  q  r 


r 


2t  .  31_7  .  7_  V-2 


(b  \     /         n  \     ac+b    mr-\-n    ac-\-b         r      _  acr-\-br 
a+c  r\  m^"r  f         c      !       r         ~c     ^mr+n~ cmr-\~cn' 


88 


RAYS  ALGEBRA,  FIRST  BOOK. 


In  like  manner,  reduce  the  following  complex  to  simple 
fractions  : 


1. 


h 

a 
3 


A         ad 

Ans.  — 

be 


Ans. 


21 

2a 


«+• 


3. 


Ans. 


Ans. 


ac-f  1 


H-r 


A  complex   fraction   may  also  be   reduced  to   a  simple 

one,  by  multiplying  both  terms  by  the  least  eommon  multifile 

of  the  denominators  of  the  fractional  parts  of  each  term. 

41 
Thus,  to  reduce  _?,  multiply  both  terms  by  6;  the  result  is  §f* 
oi  *•* 


144.  Resolution  of  fractions  into  series. 

An  Infinite  Series  is  an  unlimited  succession  of  terms, 
which  observe  the  same  law. 

The  Law  of  a  Series  is  a  relation  existing  between  its 
terms,  such  as  that  when  some  of  them  are  known  the 
others  may  be  found. 

Thus,  in  the  infinite  series,  1 — ax-\-a2x2 — a3rc3-j-a4a;4,  etc.,  any 
term  may  be  found  by  multiplying  the  preceding  term  by  — ax. 

Any  proper  algebraic  fraction  whose  denominator  is  a  polynomial 
can,  by  division,  be  resolved  into  an  infinite  series;  for  the  process 
of  division  never  can  terminate. 

After  a  few  of  the  terms  of  the  quotient  are  found,  the  law  of  the 
series  will,  in  general,  be  easily  discovered. 

Review. — 143.  How  reduce  a  complex  fraction  to  a  simple  one 
by  division?     How,  by  multiplication? 

144.  What  is  an  infinite  series?  What  is  the  law  of  a  series? 
Give  an  example,  Why  can  any  proper  algebraic  fraction  whoso 
denominator  is  a  polynomial,  be  resolved  into  an  infinite  series 
by  division? 


INFINITE  SERIES.  89 

1.  Convert  the  fraction  ■= into  an  infinite  series. 

1 — x 

\l—x 


1—x        l-f#-|-a;2-[-a;3-j-,  etc.  The  law  of  this  series  evidently 

-\X  is,  that  each  term  is  equal  to  the 

-\-X — X2  preceding  term  multiplied  by  -\-X. 

+&~ 
-fa:2— ic3 
-frc3 

From  this,  it  appears  that  the  fraction  -z is  equal  to  the  infinite 

series,  l+#+ X2-^-Xz-{-Xi-\-1  etc. 

Resolve  the  following  into  infinite  series  by  division : 
2.     1        s=cl — x-\-x2 — x3-\-x4 — ,  etc.,  to  infinity. 

CLOT  Ob  Cu  Ou 

3. -=x-\ \-—A — H-,  etc.,  to  infinity. 

a — x  a      a1     a3  J 

\ x 

4.     _— =l_2a;-f- 2a:2—  2a;3-f ,  etc.,  to  infinity. 

5-  3i=1+^-i+J-' etc" t0  infinity- 


GENERAL    REVIEW. 

Define  a  fraction.  An  entire  quantity.  A  mixed  quantity.  An 
Improper  fraction.  Simple.  Compound.  Complex.  Terms  of  a 
fraction.  Denominator.  Numerator.  State  Proposition  I.,  and  illus- 
trate it.     Proposition  II. ;   III.;  IV.;  V.;  VI. 

How  reduce  a  fraction  to  its  lowest  terms?  To  an  entire  or  mixed 
quantity?  How  a  mixed  quantity  to  a  fraction?  Rule  for  the  signs 
of  fractions.  How  reduce  to  common  denominators?  To  least  com- 
mon denominators? 

How  add  fractions?  Subtract?  Multiply?  Divide?  How  re- 
duce a  complex  to  a  simple  fraction  ?  How  resolve  a  fraction  into 
an  infinite  series? 

Define  mathematics.  Algebra.  Theorem.  Problem.  Factor. 
Coefficient.  Exponent.  Power.  Root.  Monomial.  Binomial. 
Trinomial.  Polynomial.  Residual  quantity.  Reciprocal  of  a 
quantity.  Prime  quantity.  Composite.  Quadratic  trinomial.  The 
G.C.D.     TheL.C.M. 

1st  Bk.  8 


90  RAY'S  ALGEBRA,  FIRST  BOOK. 

IV.    SIMPLE    EQUATIONS. 

DEFINITIONS    AND  ELEMENTARY   PRINCIPLES. 

1 13.  The  most  useful  part  of  Algebra  is  that  which 
relates  to  the  solution  of  problems.  This  is  performed  by 
means  of  equations. 

An  Equation  is  an  algebraic  expression,  stating  the 
equality  between  two  quantities ;  thus,  x — 3=4,  is  an 
equation,  stating  that  if  3  be  subtracted  from  x,  the  re- 
mainder will  equal  4. 

146.  Every  equation  is  composed  of  two  parts,  sepa- 
rated from  each  other  by  the  sign  of  equality. 

The  First  Member  of  an  equation  is  the  quantity  on 
the  left  of  the  sign  of  equality. 

The  Second  Member  is  the  quantity  on  the  right  of  the 
sign  of  equality. 

Each  member  is  composed  of  one  or  more  terms. 

147.  There  are  generally  two  classes  of  quantities  in 
an  equation,  the  known,  and  the  unknown. 

The  known  quantities  are  represented  either  by  num- 
bers or  the  first  letters  of  the  alphabet,  as  a,  b,  c,  etc. ; 
the  unknown  quantities,  by  the  last  letters  of  the  alphabet, 
as,  x,  y1  z,  etc. 

148.  Equations  are  divided  into  degrees,  called  first, 
second,  third,  etc. 

The  Degree  of  an  equation  depends  upon  the  highest 
power  of  the  unknown  quantity  which  it  contains. 

Preview. — 145.  What  is  an  equation  ?  Example.  146.  Of  how 
miny  parts  composed?  How  are  tiny  separated?  What  is  that  on 
th3  left  called  ?    On  the  right?    Of  what  is  each  member  composed? 

147.  How  many  classes  of  quantities  in  an  equation?  How  ara 
the  ttnown  quantities  represented?     How  the  unknown? 


SIMPLE  EQUATIONS.  91 

A  Simple  Equation,  or  an  equation  of  the  first  degree, 
is  one  which  contains  no  power  of  the  unknown  quantity 
higher  than  the  first. 

Thus,  2a?-f-5=9,  and  ax-\-b=c,  are  simple  equations,  or 
equations  of  the  first  degree. 

A  Quadratic  Equation,  or  an  equation  of  the  second 
degree,  is  one  in  which  the  highest  power  of  the  unknown 
quantity  is  a  square. 

Thus,  4x2 — 7=29,  and  ax2-{-bx=c,  are  quadratic  equa- 
tions, or  equations  of  the  second  degree. 

So,  we  have  equations  of  the  third  degree,  fourth  degree, 
etc.,  distinguished  by  the  highest  power  of  the  unknown 
quantity. 

When  any  equation  contains  more  than  one  unknown 
quantity,  its  degree  is  equal  to  the  greatest  sum  of  the 
exponents  of  the  unknown  quantities  in  any  of  its  terms. 

Thus,  xy-\~ax-\-by=c,  is  an  equation  of  the  second  degree. 
x2y-\-x2  -\-cx=a,  is  an  equation  of  the  third  degree. 

149.  An  Identical  Equation  is  one  in  which  the  two 
members  are  identical ;  as,  5=5,  or  2x — l=2x — 1. 

Equations  are  also  distinguished  as  numerical  and  literal. 

A  Numerical  Equation  is  one  in  which  all  the  known 

quantities  are  expressed  by  numbers;  as,  x2-j-2x=Sx-\-}7. 

A  Literal  Equation  is  one  in  which  the  known  quan- 
tities are  represented  by  letters,  or  by  letters  and  num- 
bers;  as,  ax — b=cx-\-d,  and  ax2-\-bx=2x—  5. 

Review. — 148.  How  are  equations  divided?  On  what  does  the 
degree  depend?  What  is  a  simple  equation,  or  an  equation  of  the 
first  degree?  Example.  What  is  a  quadratic  equation,  or  an  equa- 
tion of  the  second  degree  ?     Example. 

148.  When  an  equation  contains  more  than  one  unknown  quan- 
tity, to  what  is  its  degree  equal?  Example.  149.  What  is  an  iden- 
tical equation?  Examples.  A  numerical  equation?  Example.  A 
literal  equation  ?     Example. 


92  RAY'S  ALGEBRA,  FIRST  BOOK. 

115©.  Every  equation  is  to  be  regarded  as  the  state- 
ment, in  algebraic  language,  of  a  particular  question. 

Thus,  x — 3=4,  may  be  regarded  as  the  statement  of 
the  following  question : 

To  find  a  number,  from  which,  if  3  be  subtracted,  the 
remainder  will  be  equal  to  4. 

Adding  3  to  each  member,  we  have  x — 3-f3=4-f  3, 
or  x==l. 

An  equation  is  said  to  be  verified,  when  the  value  of  the 
unknown  quantity  being  substituted  for  it,  the  two  mem- 
bers are  rendered  equal. 

Thus,  in  the  equation  x — 3=4,  if  7,  the  value  of  x,  be 
substituted  instead  of  it,  we  have  7 — 3=4,  or  4=4. 

To  solve  an  equation  is  to  find  the  value  of  the  unknown 
quantity;  or,  to  find  a  number  which,  being  substituted 
for  the  unknown  quantity,  will  render  the  two  members 
identical. 

151.  The  Boot  of  an  equation  is  the  value  of  its  un- 
known quantity. 


SIMPLE  EQUATIONS  CONTAINING  ONE  UNKNOWN 
QUANTITY. 

152.  All  rules  for  finding  the  value  of  the  unknown 
quantity  in  an  equation  are  founded  on  this  evident  prin- 
ciple : 

If  we  perform  exactly  the  same  operation  on  two  equal 
quantities,  the  results  will  be  equal. 

This  important  principle  may  be  otherwise  expressed  by 
the  following 

Review. — 150.  How  is  every  equation  to  be  regarded?  Example. 
When  said  to  be  verified?  What  is  solving  an  equation?  151.  What 
is  the  root  of  an  equation? 

152.  Upon  what  principle  are  the  operations  used  in  solving  an 
equation  founded? 


SIMPLE  EQUATIONS.  93 

AXIOMS. 

1.  If,  to  two  equal  quantities,  the  same  quantity  be  added, 
the  sums  icill  be  equal. 

2.  If,  from  two  equal  quantities,  the  same  quantity  be  sub' 
traded,  the  remainders  will  be  equal. 

3.  If  two  equal  quantities  be  multiplied  by  the  same  quart' 
tity,  the  products  will  be  equal. 

4.  If  two  equal  quantities  be  divided  by  the  same  quan- 
tity, the  quotients  will  be  equal. 

5.  If  two  equal  quantities  be  raised  to  the  same  power, 
the  results  will  be  equal. 

6.  If  the  same  root  of  two  equal  quantities  be  extracted, 
the  results  will  be  equal. 

Remark. — An  axiom  is  a  self-evident  truth.  The  preceding 
axioms  are  the  foundation  of  a  large  portion  of  the  reasoning  in 
mathematics. 

1«>3«  There  are  two  operations  of  constant  use  in  the 
solution  of  equations,  Transposition  and  Clearing  an  Equa- 
tion of  Fractions.     These  we  are  now  to  consider. 


TRANSPOSITION. 

Suppose  we  have  the  equation  x — 3=5. 

Since,  by  Art.  152,  the  equality  will  not  be  affected  by 
adding  the  same  quantity  to  both  members,  if  we  add  3  to 
each  member,  we  have  x — 3+3=5+3. 

But,  — 3+3=0  ;  omitting  these,  we  have  #=5+3. 

Now,  the  result  is  the  same  as  if  we  had  transposed  the 
term  — 3,  to  the  opposite  member  of  the  equation,  and,  at 
the  same  time,  changed  its  sign. 

Again,  take  the  equation  x+b=c. 

Review. — 152.  What  are  the  axioms  which  this  principle  em- 
braces? 153.  What  two  operations  arc  constantly  used  in  the  solu- 
tion of  equations  ? 


94  BAYS  ALGEBRA,  FIRST  BOOK. 

If  we  subtract  b  from  each  side,  Art.  152,  Axiom  2,  we 
have  x-\-b — b=c — b,  or  x=c — b. 

But,  this  result  is  also  the  same  as  if  we  had  transposed 
the  term  -\-b,  to  the  opposite  side,  changing  its  sign. 
Hence, 

Any  quantity  may  be  transposed  from  one  side  of  an  equa- 
tion to  the  other,  if  at  the  same  time,  its  sign  be  changed. 


TO    CLEAR    AN    EQUATION    OF    FRACTIONS. 

1>>4. — 1.  Let  it  be  required  to  clear  the  following 
equation  of  fractions  : 

xx 

If  we  multiply  the  first  member  by  2,  the  denominator  of  the  first 
fraction  will  be  removed;  but  if  we  multiply  the  first  member 
by  2,  we  must  multiply  the  other  member  by  the  same  quantity, 
Art.  152,  Axiom  3,  in  order  to  preserve  the  equality.  Multiply- 
ing both  sides  by  2,  we  have 

In  like  manner,  multiplying  both  sides  by  3,  we  have 

3#-f2a;:=30. 

Instead  of  this,  it  is  plain  that  we  might  have  multiplied  at  once, 
by  2X3  ;   that  is,  by  the  product  of  the  denominators. 

2.  Again,  clear  the  following  equation  of  fractions : 

ab  '  6c 

Reasoning  as  before,  we  first  multiply  both  sides  by  Ctb,  and  then 
by  be,  or  at  once  by  aby(bc,  and  the  equation  will  be  cleared  of 
fractions,  and  we  shall  have  bcx-}-abx—ab2cd. 

Review. — 153.  How  may  a  quantity  be  transposed  from  one  mem- 
ber of  an  equation  to  the  other?  Explain  the  principle  by  an 
example. 


SIMPLE  EQUATIONS.  95 

Instead  of  multiplying  every  term  by  abybc,  it  is  evident  thrt 
if  each  term  be  multiplied  by  the  L.C.M.  of  the  denominators, 
which,  in  this  case,  is  abc,  the  denominators  will  be  removed; 
thus,  ex-\-ax=abcd.    Hence, 


TO    CLEAR   AN    EQUATION    OF    FRACTIONS, 

Eule. — Find  the  least  common  multiple  of  all  the  denom- 
inators, and  multiply  each  term  of  the  equation  by  it. 

Clear  the  following  equations  of  fractions : 

1.  g-f|==5 Ans.  3.t4-2^-80. 

2.  %  —  %=2 Ans.  4x—2>x=-2\. 

3  4 

3.  5+£— ?=™ Ans.  6*+  Sx— 4^_1 0. 

4  o      b      \L 

.    2x — 3  ,  x    x — 3      5 
~T~  +  j=~2T'i~U' 

Ans   14x— 21+4^=14^— 42+10. 

5.  x—  ^=5— ^.  Ans.6aj— 3a;+9=30— 2»— 8. 

6.  -+—^-=6 Ans.  2x-\-ax — da=2ab. 

a        Z 

7.  JL^tyL&  .     .    Ans.  4$+8ax— 24a=9x—27. 
x — 6      6      4 

x+1     3— c 

8.   s+ r=«- 

cc — o     a — b 

Ans.  ax — bx-\-  a — b-\-  Sx — ex — 9  +  3c=a2x — abx — 3a2 -f  Sab. 

9.  — r^"4" r——> — 7T,.       •  Ans.  ax — bx-\-ax-\-bx:=c. 

a-\-b     a — b     a1 — bz 

10.   -^ +-£-{- jL=fc.     .     .    Ans.  adf+bcf+bde=:bJfhx. 
ox     ax    jx 

Review. — 154.  How  clear  an  equation  of  fractions?     Explain  the 
principles  by  the  examples  given. 


96  RAY'S  ALGEBRA,  FIRST  BOOK. 

SOLUTION    OF    SIMPLE    EQUATIONS,    CONTAINING 
ONE    UNKNOWN    QUANTITY. 

155.  The  unknown  quantity  in  an  equation  may  be 
combined  with  the  known  quantities,  by  addition,  sub- 
traction, multiplication,  or  division ;  or,  in  two  or  more  of 
these  methods. 

1.  Let  it  be  required  to  find  the  value  of  x,  in  the 
equation  a?-}- 3=5,  where  the  unknown  quantity  is  con- 
nected by  addition. 

By  subtracting  3  from  each  side,,  we  have  cc=5 — 3=2. 

2.  Let  it  be  required  to  find  the  value  of  x,  in  the 
equation  x — 3=5,  where  the  unknown  quantity  is  con- 
nected by  subtraction. 

By  adding  3  to  each  side,  we  have  #=5 +  3=8. 

3.  Let  it  be  required  to  find  the  value  of  x,  in  the 
equation  3a*=15,  where  the  unknown  quantity  is  connected 
by  multiplication. 

By  dividing  each  side  by  3,  we  have  cc=-=-=5. 

o 

4.  Let  it  be   required  to   find  the  value   of  x,  in  the 

equation  Q=2,  where  the  unknown  quantity  is  connected 
o 

by  division. 

By  multiplying  each  side  by  3,  we  have  #=2x3=6. 

Prom  the  solution  of  these  examples,  we  see  that 

When  the  unknown  quantity  is  connected  by  addition,  it  is 
to  be  separated  by  subtraction. 

When  connected  by  subtraction,  it  is  separated  by  addition. 
When  connected  by  multiplication,  it  is  separated  by  division. 
When  connected  by  division,  it  is  separated  by  multiplication. 


SIMPLE  EQUATIONS.  97 

5.  Find  the  value  of  cc,  in  the  equation  Zx — 3=zx-\-&. 
By  transposing  the  terms  — 3  and  x,  we  have 

3x—x=5+3 
Reducing,         2#=8 
Dividing  by  2,  a;=|=4. 

Let  4  be  substituted  for  x,  in  the  original  equation,  and,  if  it  is 
the  true  value,  it  will  render  the  two  members  equal. 

Original  equation,     .     .     .     3# — 3=rc-|-5. 
Substituting  4  in  the  place  of  x,  it  becomes 
3x4—3=4+5,  or  9=9. 

This  method  of  substituting  the  value  of  the  unknown 
quantity  instead  of  itself,  is  called  verification. 

x 2  rc+2 

6.  Find  the  value  of  x  in  the  equation  x ^— =4+— -=— 

o  5    * 

Multiplying  both  sides  by  15,  the  L.C.M.  of  the  denominators,  we 
have  lhx—  (5x— 10)=60+3a:-f  6; 

or,  15a;—  5#+10  — 60+3a;-f  6. 
By  transposition,  15a: —  5x — Sx  =60-f  6  — 10. 
Reducing,  7#=56. 

Dividing,  x=8. 

7.  Find  the  value  of  x  in  the  equation  T  — d= — \-c 

b  a 

Multiplying  both  sides  by  ab,     ax—abd=bx  A-abc. 
Transposing,  ax — bx  =abc-\-abd. 

Separating  into  factors,  (a — b)x=ab(c-\-d). 

Dividing  by  (a — 6),  x= — ' — ~-\ 

From  the  preceding  examples  and  illustrations,  we  de- 
rive the 

Review. — 155.  How  may  the  unknown  quantity  in  an  equation 
be  combined  with  known  quantities?     Examples. 

155.  When  the  unknown  quantity  is  connected  by  addition,  how 
is  it  to  be  separated?     When,  by  subtraction?     By  multiplication? 
By  division?     What  is  verification? 
1st  Bk.  9* 


98  RAYS  ALGEBRA,  FIRST  BOOK. 

RULE, 
FOR   THE    SOLUTION   OP   SIMPLE    EQUATIONS. 

1.  If  necessary,  clear  the  equation  of  fractions. 

2.  Transpose  all  the  terms  containing  the  unknown  quan- 
tity to  one  side,  and  the  known  quantities  to  the  other. 

3.  Combine  the    terms  in    each   member   by  the  rule  for 
addition. 

4.  Divide  both  members  by  the  coefficient  of  the  unknown 
quantity. 


EXAMPLES    FOR    PRACTICE. 

Note. — Verify  the  value  of  the  unknown  quantity  in  each  ex- 
ample. 


1.  3x— 5=2*4- 1 

2.  3aj— 8=16— 5aj 

3.  Sx— 25=— x— 9 

4.  15— 2x=6x— 25 

5.  b(x+l)+6(x+2)=9(x+S). 


Ans.  #=12. 
Ans.  x=S, 
Ans.  ic=4. 
Ans.  x=h. 
Ans.  x=b. 


6.  10(aj+5)  +  8(*+4)=5(a;+13)+121.    Ans.  <c=8. 

.Ans.  #=10. 


*  l-^-l 


(-.      X   .    X  X       _,    i 

8-2+3-4=14' 
3*      2#        X 


10. 
11 


x— 2 


4 
3.x-f-l 


2=1 

2x 


#4-7 


10- 


12.  Sx~2      4-#^ 


1x-2 


.AnS.    #=24. 

.  Ans.  #=2. 
.  Ans.  x=2. 
.  Ans.  #=14. 
.    Ans.  #=2. 


Review. — 155.  What  is  the  rule  for  the  solution  of  an  equation 
/>f  the  first  degree  containing  one  unknown  quantity? 


SIMPLE  EQUATIONS. 


13.  i*—  ja+18=g(4*+l). 


14. 


5a; 


x+4~ 


1. 


15    2x    x-2-x+x+1S 
lb.  Zx—  1Q  =*f    15    . 

3     ,_2_5     a;+3 
4 


16. 


4 

17.  2x 


3   — 4        4   '    * 

2*+ 11     4*— 6        7—8; 


11 


1Q   cc+7     ,3    2^+5,10— hx 
18.  -g 51=-^-+-^—. 


19. 


2Q- !)_>>— 4     »— 1 
60  ' 


8  '        5         "15 

20.  4x—b=2x—d.     .    . 

21.  ax-\-b=cx-{-d.      .     . 

22.  ax — bx=c-\rdx — e.    . 

23.  7-f9a— bx=6x+bax. 


99 

.  Ans.  cc=20. 
.    Ans.  a5=l. 


Ans.  jc=li. 


Ans.  8=3.1. 
Ans.  sc===7. 
Ans.  cc=8. 


Ans.  a* 
Ans.  a: 
Ans.  x- 
Ans.  a; 


Ans.  a:=2. 
b—d 


—   2   " 
_tf— b 

a — c 
c — e 


a — b — d' 

9a+7 


24.  (a+b)(b— *)  +  («— b)(a+x)=c\  Ans.  a:: 


5a+ll' 

a2+b2—c2 


2b 


25.  --4-T=C AnS.    X= rr, 

a     6  a-J-o 

26.  -4-^4--=l Ans.  x=a+b+c. 

^    x  .  x  ,  x      1  .  a&cd 

27.  -+=-4--=** Ans.  s= 

a     b     c  ab-\-ac-\-bc 

28.  ^+?_c+l-=2.    .    . 

a;      a;      a; 


.  Ans.  a;=^(a&-}-ac+6c). 


29.  ?+---=0. 

a;     c      e 


Ans.  a;: 


cc£ — be 

1 


30.  J±?=l+i Ans.  x--{ 

1—x  a  2a+l 


100  KAY'S  ALGEBRA,  FIRST  LOOK. 

31.  — =ab-\-b-\ — Ans.  x= — ; — 

xx  b 

-wj.    a — b a-\-b  Sac — be 

x — c     x-\-2c  26 


QUESTIONS  PRODUCING  SIMPLE  EQUATIONS,  CONTAINING 
ONE  UNKNOWN  QUANTITY. 

156.  The  solution  of  a  problem,  by  Algebra,  consists 
of  two  distinct  parts  : 

1.  Expressing  the  conditions  of  the  problem  in  algebraic 
language;  that  is,  forming  the  equation. 

2.  Solving  the  equation,  or  finding  the  value  of  the  unknown 
quantity. 

The  first  is  the  most  difficult  part  of  the  operation. 

Sometimes,  the  statement  of  the  question  furnishes  the 
equation  directly ;  and,  sometimes,  it  is  necessary,  from 
the  conditions  given,  to  deduce  others  from  which  to  form 
the  equation.  In  the  one  case,  the  conditions  are  called 
explicit  conditions  ;  in  the  other,  implied  conditions. 

It  is  impossible  to  give  a  precise  rule  for  forming  an 
equation.  The  first  point  is  to  understand  fully  the  nature 
of  the  question  or  problem. 

After  this,  the  equation  may  generally  be  formed  thus : 

Rule. — Denote  the  required  quantity  by  one  of  the  final 
letters  of  the  alphabet;  then,  by  means  of  signs,  indicate  the 
same  operations  that  it  would  be  necessary  to  make  on  the 
answer  to  verify  it. 

Review 156.  Of  what  two  parts  does  the  solution  of  a  problem 

consist?     What  are  explicit  conditions?     Implied  conditions? 

156.  By  what  rule  may  the  equation  of  a  problem  generally  be 
formed  ? 


SIMPLE  EQUATIONS.  101 


EXAMPLES. 

1.  There  are  two  numbers,  the  second  of  which  is  three 
times  the  first,  and  their  sum  is  48 ;  what  are  the  num- 
bers ? 

Let  x—  the  first  number. 

Then,  by  the  first  condition,  3x=  the  second. 

And,  by  the  second  condition,  #-j-3:c=48. 

Reducing,  4x=48. 

Dividing  by  4,  a;=12,  the  smaller  number. 

Then,  3#=36,  the  larger  number. 

Proof,  or  verification.     12-J-36— 48. 

2.  A  father  said  to  his  son,  "  The  difference  of  our  ages 
is  48  years,  and  I  am  5  times  as  old  as  you."  What  were 
their  ages  ? 

Let  x=  the  son's  age. 
Then,  5x=  the  father's  age. 
And  5x — #=48. 
Reducing,  4#=48. 
Dividing,  #=12,  the  son's  age. 
Then,  5#=60,  the  father's  age. 

Verification.     60 — 12=48,  the  difference  of  their  ages. 

3.  What  number  is  that,  to  which  if  its  third  part  be 
added,  the  sum  will  be  16? 

Let  x—  the  required  number. 

Then,  the  third  part  of  it  will  be  represented  by  s 

And,  by  the  conditions  of  the  question,  we  have  the  equation 
*+*=16. 

Multiplying  by  3,  to  clear  it  of  fractions,  3x-\-x—48. 
Reducing,  4#=48;  and  dividing,  #=12. 

Verification.     12-f-i^=12-f  4=16. 

Note. — The  pupil  should  verify  the  answer  in  every  example. 


102  RAYS  ALGEBRA,  FIRST  BOOK. 

4.  What  number  is  that,  which  being  increased  by  its 
half,  and  then  diminished  by  its  two  thirds,  the  remainder 
will  be  105? 


Let  X==  the  number.  „ 

Then,  the  one  half  equals  «,  and  the  two  thirds,  -s-. 

x      t2x 
And,  by  the  question,  x-\-~ — «-=105. 

Multiplying  by  6,  6z-f  Sx— 4z= 630. 
Reducing,  5£=630;  and  dividing,  £-=126. 

It  is  sometimes  better  to  simply  indicate  the  multiplication,  thus: 

x     2x 
*+2-lf=105 
Qx^Sx— 4,Tz=105x6 
5^=105x6 
x=  21x6=126. 

5.  It  is  required  to  divide  a  line  25  inches  long,  into 
two  parts,  so  that  the  greater  shall  be  3  inches  longer  than 
the  less. 

Let  X—  the  length  of  the  smaller  part. 

Then,  x+3=  the  greater  part. 

And  by  the  question,  £+£+3=25. 

Reducing,  2£+ 3=25. 

Transposing  3,  2£=25— 3=22. 

Dividing,  £=11,  the  smaller  part;  and  £+3=14,  the  greater. 


6.  It  is  required  to  divide  $68  between  A,  B,  and  C,  so 
that  B  shall  have  $5  more  than  A,  and  C  $7  more  than  B. 

Let£=  A's  share.     Then,  £+5=  B's  share;  and  £+12=  C's. 
Then,  by  the  terms  of  the  question,  £+(£+5)+(£+12)=68. 
Reducing,  3£+ 17=68. 
Transposing,  3£=68— 17=51. 
Dividing,  £=17,  A's  share. 

£+  5=22,  B's  share. 

£+12=29,  C's  share. 


SIMPLE  EQUATIONS.  103 

*T.  What  number  is  that,  which  being  added  to  its  third 
part,  the  sum  will  be  equal  to  its  half  added  to  10? 

Let  x  represent  the  number. 

'  x 

Then,  the  number,  with  its  third  part,  is  represented  by  x-\- ^ ; 

x 
and  its  half,  added  to  10,  is  expressed  by  ;r-|-10.     By  the  condi- 

xx 
tions  of  the  question,  these  are  equal;   that  is,  x-\- -=--J-10. 

Multiplying  by  6,  6x+2x=3x  +  60. 

Iteducing  and  transposing,     8x — 3a;=b0. 

5z=60. 
Dividing,      a;— 12. 
Verification.     12+  1g=r  ^+10 ;  or,  16=16. 

Hereafter,  we  shall  omit  the  terms,  transposing,  dividing,  etc.,  as 
the  steps  of  the  solution  will  be  evident  by  inspection. 

8.  A  cistern  was  found  to  be  one  third  full  of  water, 
and,  after  emptying  into  it  17  barrels  more,  it  was  found 
to  be  half  full ;  what  number  of  barrels  will  it  contain 
when  full  ? 

Let  X=  the  number  of  barrels  the  cistern  will  contain. 

r±hen,g+l/=£. 

£c+102=3# 

102=#;  or,  transposing,  — x— — 102. 
And  multiplying  both  sides  by  — 1,  we  have  #=102. 
It  is  most  convenient  to  make  the  unknown  quantity  stand  on  the 
left   side  of  the   sign  of  equality.     If   negative,   the  sign   may  be 
changed  as  above  shown.     Or,  in  general, 

The  signs  of  all  the  terms  of  loth  members  of  an  equa- 
tion may  be  changed  at  pleasure,  since  this  would  be  the 
result  of  multiplying  by  — 1. 

9.  A  cistern  is  supplied  with  water  by  two  pipes  ;  the 
less  alone  can  fill  it  in  40  minutes,  arid  the  greater  in  30 
min. ;  in  what  time  will  they  fill  it,  both  running  at  once  ? 

Let  x=  the  number  of  min.  in  which  both  together  can  fill  it 
Then,  -=  the  part  which  both  can  fill  in  1  min. 


104  RAY'S  ALGEBRA,  FIRST  BOOK. 

Since  the  less  can  fill  it  in  40  min.,  it  fills  JU  of  it  in  1  min. 
Since  the  greater  can  fill  it  in  30  min.,  it  fills  ^  of  it  in  1  min. 
Hence,  the  part  of  the  cistern  which  both  can  fill  in  1  min.,  is  rep- 
resented by  jTv-f  o7v>  ant*  also>  ky  -• 

ill 

Hence'  40+30=* 

Multiply  both  sides  by  120a:,  and  we  have  8#-f4a:— 120. 

7:r=:120. 
x=i §°=17^  min. 

10.  A  can  perform  a  piece  of  work  in  5  days,  B  in  6 
days,  and  C  in  8  days ;  in  what  time  can  the  three  per- 
form it  ? 

Let  X—  the  number  of  days  in  which  all  three  can  do  it. 

Then,  -=  the  part  which  all  can  do  in  1  day. 

If  A  can  do  it  in  5  days,  he  does  1  of  it  in  1  day. 

If  B       "         "6       "         "         I      "         " 

If  C       "        "8      "        "        I     "        " 

Hence,  the  part  of  the  work  done  by  A,  B,  and  C  in  1  day,  is 

,,111,  1  ,r  1111 

represented  by  g+g+gi  also  by  -.     Hence,  ^+ --f  -=-. 

Or,  24a:-f20a:-|-15a;=120. 
59a;=120 

11.  How  many  pounds  of  sugar  at  5  cents  and  at  9  cents 
per  pound,  must  be  mixed,  to  make  a  box  of  100  pounds, 
at  6  cents  per  pound  ? 

Let  x=  the  number  of  pounds  at  5  cents. 
Then,  100 — X=  the  number  of  pounds  at  9  cents. 
Also,  5x=  the  value  of  the  former  in  cents. 
And  9(100 — x)-=  the  value  of  the  latter  in  cents. 
And  600=  the  value  of  the  mixture  in  cents. 
But  the  value  of  the  two  kinds  must  be  equal  to  that  of  the  mix- 
ture.    Therefore,  &r-f  9(100— £)=600 
5:r-j-900~9rc=600 
— 4z=— 300 

#=75,  the  number  of  lbs.  at  5  cts. 
100— x=25,       "         "  "  9  cts. 


SIMPLE  EQUATIONS.  105 

12.  A  laborer  was  engaged  for  30  days.  For  each  day 
he  worked,  he  received  25  cents  and  his  board  ;  and,  for 
each  day  he  was  idle,  he  paid  20  cents  for  his  board.  At 
the  expiration  of  the  time,  he  received  $3  ;  how  many  days 
did  he  work,  and  how  many  was  he  idle  ? 

Let  x=  the  number  of  days  he  worked. 
Then,  30—  X=  the  number  of  days  he  was  idle. 
Also,  25a;=  wages  due  for  work. 

And  20(30 — sc)=  the  amount  to  be  deducted  for  boarding. 
Therefore,  25#— 20(30— z)=300 
25a;— 600-f-20x  =300 
45a;=900 

a?=20,  the  number  of  days  he  worked. 
30 — #=10,  the  number  of  days  he  was  idle. 

Proof.     25x20=500  cents,  =  wages. 

20X10=200  cents,  =  boarding. 

300  cents,  =  the  remainder. 

In  the  above  example,  we  reduce  the  $3  to  cents  ;  for  it  is  evident 
we  can  add  and  subtract  only  quantities  of  the  same  denomina- 
tion. And,  since  we  can  compare  only  quantities  of  the  same 
name,  therefore,  , 

All  the  quantifies  in  both  members  of  an  equation  must 
be  of  the  same  denomination. 


13.  A  hare  is  50  leaps  before  a  greyhound,  and  takes 
4  leaps  to  the  greyhound's  3  ;  but  2  of  the  greyhound's 
leaps  are  equal  to  3  of  the  hare's ;  how  many  leaps  must 
the  greyhound  take  to  catch  the  hare? 

Let  X  be  the  number  of  leaps  taken  by  the  hound.     Then,  since 

the  hare  takes  4  leaps  while  the  hound  takes  3,  the  number  taken 

Ax 
by  the  hare,  after  the  hound  starts,  will  be  -^  ;  and  the  whole  num- 

Ax     ** 

ber  of  leaps  taken  by  the  hare  will  be  -Q--{-50,  which  is  equal,  in  ex- 

o 

tent,  to  the  x  leaps  run  by  the  hound.     But  2  leaps  of  the  hound 


106  RAYS  ALGEBRA,  FIRST  BOOK. 

are  =  3  of  the  hare's,  or  1  leap  =|  leaps  of  the  hare ;  hence,  x  leaps 
of  the  hound  =  -^   leaps  of  the  hare;  and  we  have  the  equation 

9z=8a;-f-300 
#=300,  leaps  taken  by  the  greyhound. 

14.  The  hour  and  minute  hands  of  a  watch  are  exactly 
together  between  8  and  9  o'clock ;  required  the  time. 

Let  the  number  of  min.  more  than  40  be  denoted  by  x;  that  is, 
let#=  the  min.  from  VIII  to  the  point  of  coincidence,  P;  then,  the 
hour  hand  moves  from  VIII  to  the  point  P,  while  the  min.  hand 
moves  from  XII  to  the  same  point;  or,  the  former  moves  over  a:  min., 
while  the  latter  moves  over  40-j-#  min.;  but  the  min.  hand  moves 
12  times  as  fast  as  the  hour  hand. 
Therefore,  12x=40+a; 
lla;=40 

#=40-  min.  =3  min.,  38-JL  sec.     Hence, 

The  required  time  is  43  min.  38^.  sec.  after  8  o'clock. 

15.  A  person  spent  one  fourth  of  his  money,  and  then 
received  $5.  He  next  spent  one  half  of  what  he  then 
had,  and  found  that  he  had  only  $7  remaining ;  what  sum 
had  he  at  first? 

Let  x=  the  number  of  dollars  he  h*d, 

x 

-=  what  he  spent  the  first  timj. 

4  r  Sx 

Subtracting  and  adding  5,   -^-\-&~  what  he  had  left. 

3x     5 

One  half  of  this,  -^-j-^=  what  he  ^pent  the  second  time. 

3x     5 

Subtracting  from  above,    i;  4"»ass  wnat  ne  had  left>  second  tinie- 

By  the  conditions  — — }-  ==  7 

3#-f-20=  56 

3x=  36,  and  #=12.  Ans. 

16.  Divide  42  cents  between  A  and  B,  giving  to  B  twice 
as  many  as  to  A.  Ans.  A  14,  B  28. 


SIMPLE  EQUATIONS.  107 

17.  Divide  48  into  three  parts,  so  that  the  second  may 
he  twice,  and  the  third  three  times  the  first. 

Ans.  8,  16,  and  24. 

18.  Divide  60  into  3  parts,  so  that  the  second  may 
be  3  times  the  first,  and  the  third  double  the  second. 

Ans.  6,  18,  and  36. 

19.  A  boy  bought  an  equal  number  of  apples,  lemons, 
and  oranges  for  56  cents ;  for  the  apples  he  gave  1  cent, 
for  the  lemons  2  cents,  and  for  the  oranges  5  cents  apiece ; 
how  many  of  each  did  he  purchase  ?  Ans.  7. 

20.  Bought  5  apples  and  3  lemons  for  22  cents  ;  gave 
as  much  for  1  lemon  as  for  2  apples  ;  what  did  I  give  for 
each  ?  Ans.  2  cents  for  an  apple,  4  for  a  lemon. 

21.  A's  age  is  double  that  of  B  ;  the  age  of  B  is  twice 
that  of  C;  the  sum  of  their  ages  is  98  years  ;  what  is  the 
age  of  each?  Ans.  A  56,  B  28,  C  14  years. 

22.  A,  B,  C,  and  D  have  among  them,  44  cents;  A 
lias  a  certain  number,  B  three  times  as  many  as  A,  C 
as  many  as  A  and  one  third  as  many  as  B,  and  D  as 
many  as  B  and  C ;  how  many  has  each  ? 

Ans.  A  4,  B  12,  C  8,  and  D  20. 

23.  Divide  55  into  two  parts,  in  proportion  to  each  other 
as  2  to  3. 

Let  2x=  one  part;  then,  3x=  the  other,  since  2x  is  to  Sx  as  2  is 
to  3.  2z-f-3a;=55 

5x=55 

x=U 

2z=22 


3*=33  '  AnS' 


1 

Or,  thus:  Let  x=  one  part;  then,  55 — x=  the  other. 

By  the  question,  x  :  55 — x  :  :-  2  :  3.      Then,  since,  in  every  pro- 
portion, the  product  of  the  means  is  equal  to  the  product  of  the 
extremes,  we  have    3z=2(55 — #)=110 — 2x. 
5^=110 
#=22,  and  55—2=33,  as  before. 


108  HAY'S  ALGEBRA,  FIRST  BOOK 

Sx 
Or,  thus :  Let  X=  one  part;  then,  — -  =  the  other. 

a     ,      i  3x      r-  2 

And  x-\ — n~=5o. 

2x+3x=ll0,  from  which  z=22,  and  -£-=33. 

The  first  method  avoids  fractions,  and  is  of  such  frequent  appli  • 
cation,  that  we  may  give  this  general  direction : 

When  two  or  more  unknown  quantities  in  any  problem  are 
to  each  other  in  a  given  ratio,  assume  each  of  them  a  mul- 
tiple of  an  unknown  quantity,  so  that  they  shall  have  to  each 
other  the  given  ratio. 

24.  The  sum  of  two  numbers  is  60,  and  the  less  is  to 
the  greater  as  5  to  7  ;  what  are  the  numbers  ? 

Ans.  25  and  35. 

25.  Divide  60  into  3  parts,  which  shall  be  in  propor- 
tion to  each  other  as  2,  3,  and  5.         Ans.  12,  18,  30. 

26.  Divide  60  into  3  such  parts,  that  J  of  the  first,  i  of 
the  second,  and  i  of  the  third  shall  be  equal. 

Ans.  12,  18,  30. 
Let  2x,  Sx,  and  ox  represent  the  parts. 

27.  What  number  is  that  whose  i,  J,  and  1  part  are 
together  equal  to  65  ?  Ans.  60. 

28.  What  number  is  that,  i  of  which  is  greater  than  \ 
by  4?  Ans.  70. 

29.  The  age  of  B  is  2i  times  that  of  A,  and  the  sum 
of  their  ages  is  76  yr.  ;  what  the  age  of  each? 

Ans.  A  20,  B  56  yr. 

30.  Divide  $440  between  A,  B,  and  C,  so  that  the  share 
of  A  may  be  {  that  of  B,  and  the  share  of  B  |  that  of  C. 

Ans.  A's  $90,  B's  $150,  C's  $200. 

31.  Four  towns  are  situated  in  the  order  of  the  let- 
ters A,  B,  C,  D.  From  A  to  D  is  120  mi. ;  from  A  to  B 
is  to  the  distance  from  B  to  C  as  3  to  5  ;  and  one  third 
of  the  distance  from  A  to  B,  added  to  the  distance  from 


SIMPLE  EQUATIONS.  109 

B  to  C,  is  three  times  the  distance  from  C  to  D  ;  how  far 
are  the  towns  apart? 
Ans.  A  to  B,  36  mi. ;  B  to  C,  60  mi. ;  C  to  D,  24  mi. 

32.  A  merchant  having  engaged  in  trade  with  a  certain 
capital,  lost  J  of  it  the  1st  year;  the  2d  year  he  gained  a 
sum  equal  to  §  of  what  remained  at  the  close  of  the  1st 
year ;  the  3d  year  he  lost  -j  of  what  he  had  at  the  close 
of  the  2d  year,  when  he  was  worth  $1236.  What  was 
his  original  capital?  Ans.  $1545. 

33.  The  rent  of  a  house  this  year  is  greater  by  5%  than 
it  was  last  year;  this  year  the  rent  is  $168  :  what  was  it 
last  year?  Ans.  $160. 

34.  Divide  the  number  32  into  2  parts,  so  that  the 
greater  shall  exceed  the  less  by  6.         Ans.  13  and  19. 

35.  At  an  election,  the  number  of  votes  given  for  two 
candidates  was  256  ;  the  successful  candidate  had  a  ma- 
jority of  50  votes  ;  how  many  votes  had  each? 

Ans.  153  and  103. 

36.  Divide  $1520  among  A,  B,  and  C,  so  that  B  may  re- 
ceive $100  more  than  A,  and  C  $270  more  than  B;  what 
is  the  share  of  each?         Ans.  A  $350,  B  $450,  C  $720. 

37.  A  company  of  90  persons  consists  of  men,  women, 
and  children  ;  the  men  are  4  more  than  the  women,  and  the 
children  10  more  than  both  men  and  women  ;  how  many 
of  each?  Ans.  18  women,  22  men,  50  children. 

38.  After  cutting  off  a  certain  quantity  of  cloth  from  a 
piece  of  45  yards,  there  remained  9  yards  less  than  had 
been  cut  off;  how  many  yards  had  been  cut  off?  Ans.  27. 

39.  What  number  is  that,  which  being  multiplied  by  7, 
gives  a  product  as  much  greater  than  20  as  the  number 
itself  is  less  than  20  ?  Ans.  5. 

40.  A  person  dying  left  an  estate  of  $6500,  to  be 
divided  among  his  widow,  2  sons,  and  3  daughters,  so 
that  each  son  shall  receive  twice  as  much  as  a  daughter, 


110  RAYS  ALGEBRA,  FIRST  COOK. 

and  the  widow  $500  less  than  all  her  children  together; 

required    the   share   of  the  widow,   and   of  each   son  and 

daughter. 

Ans.  Widow  $3000,  each  son  $1000,  each  daughter  $500. 

41.  Two  men  set  out  at  the  same  time,  one  from  London, 
the  other  from  Edinburgh;  one  goes  20,  the  other  30  miles 
a  day;  in  how  many  days  will  they  meet,  the  distance 
being  400  miles?  Ans.  8  days. 

42.  A  and  B  depart  from  the  same  place,  to  go  in  the 
same  direction ;  B  travels  at  the  rate  of  3,  and  A  at  the 
rate  of  5  mi.  an  hr.,  but  B  has  10  hr.  the  start  of  A  ;  in 
how  many  hr.  will  A  overtake  B  ?  Ans.  15  hr. 

43.  Being  asked  the  time  of  day,  I  replied,  "  If,  to  the 
time  past  noon,  there  be  added  its  1,  J,  and  |,  the  sum 
will  =  i  of  the  time  to  midnight ;"  required  the  hour. 

Ans.  50  min.  P.  M. 

44.  Divide  120  into  two  such  parts  that  the  less  may  be 
contained  in  the  greater  1\  times.  Ans.  48  and  72. 

45.  If  I  multiply  a  certain  number  by  7,  add  3  to  the 
product,  divide  this  by  2,  and  subtract  4  from  the  quotient, 
the  remainder  is  15.     What  is  the  number?       Ans.  5. 

46.  What  number  is  that,  which,  if  you  multiply  it 
by  5,  subtract  24  from  the  product,  divide  the  remainder 
by  6,  and  add  13  to  the  quotient,  will  give  the  number 
itself?  Ans.  54. 

47.  A  and  B  engaged  in  trade,  the  capital  of  B  being 
|  that  of  A;  B  gained,  and  A  lost,  $100;  after  which, 
if  |  of  what  A  had  left  be  subtracted  from  what  B  has, 
the  remainder  will  be  $134;  what  capital  had  each  at 
first?  Ans.  A  $786,  B  $524. 

48.  A  man  having  spent  $3  more  than  -~  of  his  money, 
had  $7  more  than  J  of  it  left;  how  much  had  he  at 
first?  Ans.  $75. 

49.  A  and  B  have  the  same  annual  income ;  A  saves 
\  of  his,  but  B  spends  $25  per  annum  more  than  A,  and 


SIMPLE  EQUATIONS.  Ill 

at  the  end  of  5  years  finds  he  has  saved  $200  ;  what  is 
the  income  of  each?  Ans.  $325. 

50.  In  a  quantity  of  gunpowder,  ~  of  the  whole,  plus 
10  lb.,  was  niter  ;  rfj  of  the  whole,  plus  1  lb.,  was  sulphur  ; 
and  ^  of  the  whole,  minus  17  lb.,  was  charcoal;  how  many 
lb.  of  gunpowder  were  there?  Ans.  69  lb. 

51.  Bought  a  chaise,  horse,  and  harness  for  $245  ;  the 
horse  cost  3  times  as  much  as  the  harness,  and  the  chaise 
$19  less  than  2^  times  as  much  as  both  horse  and  harness; 
what  the  cost  of  each  ? 

Ans.  Harness  $18,  horse  $54,  chaise  $173. 

52.  What  two  numbers  are  as  3  to  4,  to  each  of  which, 
if  4  be  added,  the  sums  will  be  as  5  to  6  ?  Ans.  6  and  8. 

53.  The  ages  of  two  brothers  are  now  25  and  30  years, 
or  as  5  to  6  ;  in  how  many  years  will  they  be  as  8  to  9  ? 

Ans.  15  yr. 

54.  A  cistern  has  3  pipes  ;  by  the  1st  it  can  be  filled 
in  lj  hr.,  by  the  2d  in  3j  hr.,  and  by  the  3d  in  5  hr.  • 
in  what  time  can  it  be  filled  by  all  running  at  once  ? 

Ans.  48  min. 

55.  Find  the  time  in  which  A,  B,  and  C  together  can 
perform  a  piece  of  work,  which  requires  7,  6,  and  9  days 
respectively,  when  done  singly.  Ans.  2|§  days. 

56.  From  a  certain  sum  I  took  one  third  part,  and  put 
in  its  stead  $50  ;  from  this  sum  I  took  the  tenth  part, 
and  put  in  its  stead  $37  ;  I  then  found  I  had  $100  ;  what 
was  the  original  sum?  Ans.  $30. 

57.  A  spent  |  of  his  salary  for  board,  ^  of  the  remainder 
for  clothes,  1  of  the  rest  for  books,  and  still  saved  $120 
per  annum;  what  was  his  salary?  Ans.  $375. 

58.  A  was  engaged  for  a  year  at  $80  and  a  suit  of 
clothes  ;  he  served  7  mon.,  and  received  for  his  wages  th*s 
clothes  and  $35  ;  what  was  the  value  of  the  clothes  ? 

Ans.  $28. 


112  RAY'S  ALGEBRA,  FIRST  BOOK. 

59.  A  man  and  his  wife  can  drink  a  keg  of  wine  in 
6  days,  and  the  man  alone  in  10  days;  how  many  days 
will  it  last  the  woman?  Ans.  15. 

60.  A  steamboat  that  can  run  15  mi.  per  hr.  with  the 
current,  and  10  mi.  per  hr.  against  it,  requires  25  hr.  to 
go  from  Cincinnati  to  Louisville,  and  return  ;  what  is  the 
distance  between  these  cities?  Ans.  150  mi. 

61.  In  a  mixture  of  wine  and  water,  ^  the  whole, 
plus  25  gal.,  was  wine,  and  ^  of  the  whole,  minus  5  gal., 
was  water ;  required  the  quantity  of  each  in  the  mixture. 

Ans.  85  gal.  wine,  35  gal.  water. 

62.  Required  to  divide  72  into  4  such  parts,  that  if 
the  1st  be  increased  by  2,  the  2d  diminished  by  2,  the  3d 
multiplied  by  2,  and  the  4th  divided  by  2,  the  sum,  dif- 
ference, product,  and  quotient  shall  be  equal. 

Ans.  14,  18,  8,  32. 

Let  the  four  parts  be  represented  by  x — 2,  x-{-2,  %x,  and  2x. 

63.  A  merchant  having  cut  19  yd.  from  each  of  3  equal 
pieces  of  silk,  and  17  from  another  of  the  same  length, 
found  that  the  remnants  taken  together  measured  142  yd.; 
what  was  the  length  of  each  piece?  Ans.  54  yd. 

64.  For  every  10  sheep  I  keep,  I  plow  an  acre  of  land, 
and  allow  1  acre  of  pasture  for  every  4  sheep ;  how  many 
sheep  can  I  keep  on  161  acres?  Ans.  460. 

65.  It  is  required  to  divide  34  into  2  such  parts,  that 
if  18  be  subtracted  from  the  greater,  and  the  less  be  sub- 
tracted from  18,  the  first  remainder  shall  be  to  the  second 
as  2  to  3.  Ans.  22  and  12. 

66.  A  person  was  desirous  of  giving  3  cents  apiece  to 
some  beggars,  but  found  that  he  had  not  money  enough 
by  8  cents  ;  he  therefore  gave  each  of  them  2  cents,  and 
then  had  3  cents  left ;  required  the  number  of  beggars. 

Ans.  11. 


SIMPLE  EQUATIONS.  113 

67.  A  could  reap  a  field  in  20  days,  but  if  B  assisted 
him  for  6  days,  be  could  reap  it  iu  16  days  ;  in  how  many 
days  could  B  reap  it  alone  ?  Ans.  30  days. 

68.  When  the  price  of  a  bu.  of  barley  wanted  but  3  cents 
to  be  to  the  price  of  a  bu.  of  oats  as  8  to  5,  nine  bu.  of 
oats  were  received  as  an  equivalent  for  4  bu.  of  barley 
and  90  cents  in  money  ;  what  was  the  price  of  a  bu.  of 
each?  Ans.  Oats  30  cts.,  barley  45  cts. 

69.  Four  places  are  situated  in  the  order  of  the  4  let- 
ters, A,  B,  C,  and  D  ;  the  distance  from  A  to  D  is  34  mi. ; 
the  distance  from  A  to  B  is  to  the  distance  from  C  to  D 
as  2  to  3  ;  and  \  the  distance  from  A  to  B,  added  to  £  the 
distance  from  C  to  D,  is  3  times  the  distance  from  B  to  C. 
Required  the  distances. 

Ans.  A  to  B  12,  B  to  C  4,  C  to  D  18  mi. 

70.  The  ingredients  of  a  loaf  of  bread  are  rice,  flour, 
and  water,  and  the  whole  weighs  15ft) ;  the  weight  of  the 
rice,  plus  51b,  is  ~  that  of  the  flour  ;  and  the  weight  of 
the  water  is  I  the  weight  of  the  flour  and  rice  together ; 
what  is  the  weight  of  each  ? 


o 


Ans.  Bice  21b,  flour  lOHb,  water  2 J  lb. 


GENERAL    REVIEW. 

What  is  an  equation?  Of  what  composed?  What  is  the  first 
member?  The  second?  How  separated?  Of  what  is  each  com- 
posed? How  many  classes  of  quantities  in  an  equation?  By  what 
represented? 

How  are  equations  divided?  Upon  what  does  the  degree  depend? 
Define  a  simple  equation.  A  quadratic.  Illustrate  each.  The  de- 
gree of  each.  Define  an  identical  equation.  Numerical  equation. 
Literal  equation.     Verification.     Root  of  an  equation. 

State  the  six  axioms.  Define  transposition.  How  clear  an  equa- 
tion of  fractions?  How  may  an  unknown  quantity  be  combined 
with  a  known?  How  separated,  when  combined  by  addition?  By 
subtraction?  Multiplication?  Division?  Rule  for  solution  of 
simple  equations. 

Define  mathematics.  Algebra.  Theorem.  Problem.  Exponent. 
Coefficient.  Factor.  Power.  Monomial.  Binomial.  Trinomial. 
Polynomial.  Residual  quantity.  Reciprocal  of  a  quantity.  Prime 
quantity.     Composite. 

1st  Bk.  10 


V     OF  THE  X 

UNIVERSITY  1 


114  RAYS  ALGEBRA,  FIRST  BOOK. 

SIMPLE     EQUATIONS, 

CONTAINING    TWO    UNKNOWN    QUANTITIES. 

157.  To  find  the  value  of  any  unknown  quantity,  we 
must  obtain  a  single  equation  containing  it,  and  known 
terms.     Hence, 

When  we  have  two  or  more  equations  containing  two  or 
more  unknown  quantities,  we  must  obtain  from  them  a  single 
equation  containing  only  one  unknown  quantity. 

The  method  of  doing  this  is  termed  elimination,  which 
may  be  defined  thus  : 

Elimination  is  the  process  of  deducing  from  two  or  more 
equations  containing  two  or  more  unknown  quantities,  a 
less  number  of  equations  containing  one  less  unknown 
quantity. 

There  are  three  methods  of  elimination. 

1st.  Elimination  by  Substitution. 
2d.  Elimination  by  Comparison. 
3d.    Elimination  by  Addition  and  Subtraction. 

158.  Elimination  by  Substitution  consists  in  finding 
the  value  of  one  of  the  unknown  quantities  in  one  of  the 
equations,  in  terms  of  the  other  unknown  quantity  and 
known  terms,  and  substituting  this,  instead  of  the  quan- 
tity, in  the  other  equation. 

To  explain  this,  suppose  we  have  the  following  equations,  in 
which  it  is  required  to  find  the  value  of  X  and  y. 

Note. — The  figures  in  the  parentheses  are  intended  to  number 
the  equations  for  reference. 

z+2?/=17.     (1.) 
2*+39/=:28.     (2.) 


SIMPLE  EQUATIONS. 


115 


By  transposing  2y  in  the  equation  (1),  we  have  X=\l — 2y.     Sub- 
stituting this  value  of  x,  instead  of  X  in  equation  (2),  we  have 
2(17-^)4-30=28; 
or,    34-4^+3^=28; 
or,    — #=28— 34 ;  or,  #=6 ; 
and    x=ll— 2y=ll— 12=5.     Hence, 


TO    ELIMINATE    BY    SUBSTITUTION, 

Rule. — 1 .  Find  an  expression  for  the  value  of  one  of  the 
unknown  quantities  in  either  equation. 

2.  Substitute  this  value  in  place  of  the  same  unknown 
quantity  in  the  other  equation;  there  will  thus  be  formed  a 
new  equation  containing  only  one  unknown  quantity. 

Note. — In  finding  an  expression  for  the  value  of  one  of  the  un- 
known quantities,  take  that  which  is  least  involved. 

Find  the  values  of  the  unknown  quantities  in  the  fol- 
lowing : 


1.  x-\-5y=38.  Ans.  x=3. 
3z+4y=:37.  y=l. 

2.  2x+4y=22.  Ans.  x=5. 
5x+7#=46.  y==3. 

3.-  3#+5ty=57.  Ans.  x=4. 

5x+3y=±1.  y=9. 

4.  4x— Sy=26.  Ans.a-=8. 

Zx— 4y=16.  y=2. 


5.  J— 2=1.       Ans.  x=2Q. 
5      4 

5*— 3^=10.  y=30. 

6.  ^—^=0.   Ans.  a=21. 

7        o 

|+|=2«.  ,=16. 


159.  Elimination  by  Comparison  consists  in  finding 
the  value  of  the  same  unknown  quantity  in  two  different 
equations,  and  then  placing  these  values  equal  to  each 
other. 

Review. — 157.  What  is  necessary  in  order  to  find  the  value  of 
any  unknown  quantity?  What,  when  we  have  two  equations,  con- 
taining two  unknown  quantities?  What  is  elimination?  How  many 
methods?     158.  Define  elimination  by  substitution.     Rule. 


11G 


RAY'S  ALGEBRA,  FIRST  BOOK. 


To  illustiate  this  method,  we  will  take  the  same  equations  whick 
were  used  to  explain  elimination  by  substitution. 


28— 3y 


*+2*/=17     (1.) 
2z+3#=28    (2.) 

From  equation  (1),  X—ll—2y.    From  (2),  x= 

Therefore,  28~3j/^l7—  2y ; 

or,    28—  3, ij=U— 4?/;   or,  y=G. 
Then,     z=17— 2#=17— 12=5. 

Or,  the  value  of  X  may  be  found  in  like  manner,  by  first  finding 
the  values  of  £/,  and  placing  them  equal  to  each  other.     Hence, 


TO    ELIMINATE    BY    COMPARISON, 

Rule. — 1.  Find  an  expression  for  the  value  of  the  same 
unknown  quantity  in  each  of  the  given  equations. 

2.  Place  these  values  equal  to  each  other;  there  will  thus 
be  formed  a  neio  equation  containing  only  one  unknown 
quantity. 

Find  the  values  of  the  unknown  quantities  in  the  fol- 
lowing : 


.  x+Sy=l6. 

x+by=22. 

Ans.  x=*J. 
3,=3. 

4. 

«    y   -, 

4     4 

Ans 

x=\2. 

.   5x-2y=4. 
2x-y=l. 

Ans.  x=2. 

3  +  2=8' 

y=S. 

x    y    -, 

Ans.  rr=36 

5. 

Ans. 

x=4b. 

-4-^-12 
6  +  4 

y=24. 

x     y    O 
"9~5-d- 

y=10. 

6.   ^+2y-*+|=42.     . 

Ans. 

*=20. 

o        4y 
Sx-T= 

=40+f.      . 

y=12. 

SIMPLE  EQUATIONS.  117 

160.  Elimination  by  Addition  and  Subtraction  con- 
sists in  multiplying  or  dividing  two  equations,  so  as  to 
render  the  coefficient  of  one  of  the  unknown  quantities  the 
same  in  both  ;  and  then,  by  adding  or  subtracting,  to  cause 
the  term  containing  it  to  disappear. 

To  explain  this  method,  we  will  take  the  same  equations  used  to 
illustrate  elimination  by  substitution  and  comparison. 

x+2y=M     (1.) 

2z+3?/=28    (2.) 

Multiplying  equation  (1)  by  2,  we  have 

&z+4#=34     (3.) 

2a:-f3i/=28,  equation  (2)  brought  down. 

Subtracting,  y~  6 

Then,  substituting  this  value  in  (3),  2#-f  4x6=34;  and  x=5. 

When  the  terms  containing  the  unknown  quantity  to  be  eliminated 
have  contrary  signs,  it  is  necessary  to  add.  In  illustration,  take  the 
following : 

3x— 5y=  6     (1.) 
4z+3*/=37     (2.) 

Multiplying  (1)  by  3,  and  (2)  by  5,  we  have 

9x—  15y=  18 
20z+152/=185 


Adding,  29x  =203 

x  =7 

Then,  from  (1)  3X~— %/=6;  or,  y=S. 

From  this  it  will  be  seen,  that  after  making  the  coefficients  of  the 
quantity  to  be  eliminated  the  same  in  both  equations,  if  the  signs 
are  alike,  we  subtract;  if  unlike,  we  add.     Hence, 

Review. — 159.  In  what  does  elimination  by  comparison  consist? 
Rule.  160.  In  what  elimination  by  addition  and  subtraction? 
Repeat  the  rule. 


118 


RAY'S  ALGEBRA,  FIRST  BOOII. 


TO    ELIMINATE    BY   ADDITION    AND    SUBTRACTION, 

Rule. — 1.  Multiply  or  divide  the  equations,  if  necessary, 
so  that  one  of  the  unknown  quantities  will  have  the  same 
coefficient  in  both. 

2.  Add  or  subtract  the  equations,  according  as  the  signs 
of  the  equal  terms  are  alike  or  unlike,  and  the  resulting  equa- 
tion will  contain  only  one  unknown  quantity. 

Remark. — When  the  coefficients  of  the  unknown  quantities  to 
be  eliminated  are  prime  to  each  other,  they  may  be  equalized  by 
multiplying  each  equation  by  the  coefficient  of  the  unknown  quan- 
tity in  the  other. 

If  the  equations  have  fractional  coefficients,  they  ought  to  be 
cleared  before  applying  the  rule. 

Find  the  value  of  the  unknown  quantities  in  the  fol- 
lowing : 


1.  &H-2y=21.  Ans.  x=S. 
x—2y=—l.  y=3. 

2.  3x— 2y=7.  Ans.  x=b. 
by—2x=10.  y=4. 

3.  2x— y=3.  Ans.  x=k. 
3x+2y=22.  y=5. 

4.  3*+2y=19.  Ans.  x=h. 
2x—Sy=4:.  y=2. 


b'  4  +  5 


x  ,  y 


5^3 


:9. 


6.  2~3-3- 
6  +  9-3- 


Ans.  x=20. 
y=lh. 

Ans.  #=12. 


PROBLEMS  PRODUCING  EQUATIONS  CONTAINING 
TWO    UNKNOWN    QUANTITIES. 

161.  The  problems  in  Art.  156,  were  all  capable  of 
being  solved  by  using  one  unknown  quantity.  Several  of 
them,  however,  contained  two,  and  some  more  than  two 
unknown  quantities ;  but  the  conditions  were  such  that  it 
was  easy  to  express  each  one  in  terms  of  the  other. 


SIMPLE  EQUATIONS.  119 

When  this  is  not  the  case,  it  becomes  necessary  to  use 
a  separate  symbol  for  each  unknown  quantity,  and  as  many 
equations  as  there  are  symbols. 

After  the  equations  are  obtained,  they  may  be  solved 
by  either  of  the  three  methods  of  elimination. 

Two  examples  are  given  below,  which  can  be  solved  by  using 
cither  one  or  two  unknown  quantities. 

1.  Given,  the  sum  of  two  numbers  equal  to  25,  and  their 
difference  equal  to  9,  to  find  the  numbers. 

Solution,  by  using  one  unknown  quantity. 
Let  x=  the  less  number;  then,  #-j-9=r  the  greater. 
And  x-\-x+9=25. 
2a;=16. 
x=z8,  the  less  number;  and  #-]-9=17,  the  greater. 

Solution,  by  using  two  unknown  quantities. 
Let  x=  the  greater,  and  y=  the  less. 

Then,  £-f2/=25     (1.) 

And    x— y=  9     (2.) 

Adding  (1)  and  (2),         2x=34,  and  #=17,  the  greater  number. 
Subtracting  (2)  from  (1),  2y=16,  and  £/=8,  the  less  number. 

2.  The  sum  of  two  numbers  is  44,  and  they  are  to  each 
other  as  5  to  6  ;  required  the  numbers. 

Solution,  by  using  one  unknown  quantity. 
Let  5x=  the  less  number ;  then,  Qx—  the  greater. 
And  5#+6a;=:44. 
llz=44 
#=4 
5x=20,  the  less  number. 
6a:=24,  the  greater  number. 

Solution,  by  using  two  unknown  quantities. 
Let  x=  the  less  number,  and  y=  the  greater. 

Then,  #+y=44     (1.) 

And    X  :  y  :  :  5  :  6 

Review. — 1G1.  In  solving  problems,  when  does  it  become  neces- 
sary to  use  a  separate  symbol  for  each  unknown  quantity? 


120  RAYS  ALGEBRA,  FIRST  BOOK. 

Or,  6x=5y     (2.),    by  multiplying  means  and  extremes. 
6:c+6,?/=264     (3.),    by  multiplying  equation  (1)  by  6. 
6,y=264 — by    by  subtracting  equation  (2)  from  (3). 
lly=264;  y=24,  and  #=44— 2/=20. 

Several  of  the  following  problems  may  also  be  solved  by  using 
only  one  unknown  quantity. 

3.  There  is  a  certain  number  consisting  of  two  places 
of  figures  ;  the  sum  of  the  figures  equals  6  ;  if  from  the 
double  of  the  number,  6  be  subtracted,  the  remainder  is 
a  number  whose  digits  are  those  of  the  former  in  an  in- 
verted order ;   required  the  number. 

In  solving  problems  of  this  kind,  observe  that  any  number  con- 
sisting of  two  places  of  figures  is  equal  to  10  times  the  figure  in 
the  ten's  place,  plus  the  figure  in  the  unit's  place. 

Thus,  23  is  equal  to  10x2+3.  In  a  similar  manner,  325  is  equal 
to  100x3+10x2+5. 

Let  X—  the  digit  in  the  place  of  tens,  and  y=  that  in  the  place 
of  units. 

Then,  10x-\-y=  the  number. 

And    10^+3;=:  the  number,  with  the  digits  inverted. 
Then,  x+y=6     (1.) 
And    2(10x+y)— 6=l0y+x     (2.) 
Or,     20x+2y— 6=10y+x. 
19z=8#+6 
8x= — 8^/+48,  multiplying  (1)  by  8,  and  transposing. 
27#=54,  by  adding. 

x=2,  and  y=6— 2=4.     Ans.  24. 

4.  What  two  numbers  are  those  to  which  if  5  be  added, 
the  sums  will  be  to  each  other  as  5  to  6  ;  but,  if  5  be 
subtracted  from  each,  the  remainders  will  be  to  each  other 
as  3  to  4  ? 

By  the  conditions  of  the  question,  we  have  the  following  propor- 
tions : 

z+5  :  y+5  :  :  5  :  6 
X— 5  :  y— 5  :  :  3  :  4 


SIMPLE  EQUATIONS.  121 

Since,  in  every  proportion,  the  product  of  the  means  is  equal  to 
the  product  of  the  extremes,  we  have  the  two  equations, 

4(a_5)=3(y-5) 

From  these  equations,  the  values  of  X  and  y  are  readily  found  to 
be  20  and  25. 

Note. — In  solving  the  following,  the  values  of  the  unknown 
quantities  may  be  found  by  either  method  of  elimination. 

5.  A  grocer  sold  to  one  person  5  lb.  of  coffee  and  3  lb. 
of  sugar,  for  79  cents ;  and  to  another,  at  the  same  prices, 
3  lb.  of  coffee  and  5  lb.  of  sugar,  for  73  cents  ;  what  was  the 
price  of  a  lb.  of  each?    Ans.  Coffee  11  cts.,  sugar  8  cts. 

6.  Sold  to  one  person  9  horses  and  7  cows,  for  $300; 
to  another,  at  the  same  prices,  6  horses  and  13  cows,  for 
the  same  sum  ;  what  the  price  of  each  ? 

Ans.  Horses  $24,  cows  $12. 

7.  It  is  required  to  find  two  numbers,  such  that  I  of 
the  first  and  |  of  the  second  shall  be  22,  and  j  of  the 
first  and  i  of  the  second  shall  be  12.       Ans.  24  and  30. 

8.  If  the  greater  of  two  numbers  be  added  to  \  of  the 
less,  the  sum  will  be  37  ;  but  if  the  less  be  diminished 
by  |  of  the  greater,  the  difference  will  be  20  ;  what  are 
the  numbers  ?  Ans.  28  and  27. 

9.  A  farmer  has  2  horses,  and  a  saddle  worth  $25  ;  if 
the  saddle  be  put  on  the  first  horse,  his  value  will  be  double 
that  of  the  second ;  but,  if  put  on  the  second  horse,  his 
value  will  be  three  times  that  of  the  first.  Required  the 
value  of  each  horse.  Ans.  First  $15,  second  $20. 

10.  A  and  B  are  in  trade  together;  if  $50  be  added 
to  A's  property,  and  $20  taken  from  B's,  they  will  have 
the  same  sum ;  and  if  A's  property  was  3  times,  and 
B's  5  times  as  great  as  each  is,  they  would  together  have 
$2350  ;  how  much  has  each?       Ans.  A  $250,  B  $320. 

IstBk.  11* 


122  RAY'S  ALGEBRA,  FIRST  BOOK. 

11.  A  number  consists  of  two  digits,  which,  divided  by 
their  sum,  gives  7  ;  if  the  digits  be  written  in  inverse  order, 
and  the  number  so  arising  be  divided  by  their  sum  plus  4, 
the  quotient  will  be  3.     What  the  number  ?         Ans.  84. 

12.  If  we  add  8  to  the  numerator  of  a  certain  fraction, 
its  value  becomes  2  ;  if  we  subtract  5  from  the  denominator, 
its  value  becomes  3  ;  required  the  fraction.  Ans.  |. 

13.  If  to  the  ages  of  A  and  B  18  be  added,  the  result 
will  be  double  the  age  of  A ;  but,  if  from  their  differ- 
ence 6  be  subtracted,  the  result  will  be  the  age  of  B  ;  re- 
quired their  ages.  Ans.  A  30,  B  12  yrs. 

14.  There  are  two  numbers  whose  sum  is  37,  and 
if  3  times  the  less  be  subtracted  from  4  times  the  greater, 
and  the  difference  divided  by  6,  the  quotient  will  be  6 ; 
what  are  the  numbers?  Ans.  16  and  21. 

15.  Find  a  fraction,  such  that  if  3  be  subtracted  from 
the  numerator  and  denominator,  the  value  will  be  j  ;  and 
if  5  be  added  to  the  numerator  and  denominator,  the  value 
will  be  I.  Ans.  T7¥. 

16.  A  father  gave  his  two  sons,  A  and  B,  together 
$2400,  to  engage  in  trade  ;  at  the  close  of  the  year,  A 
has  lost  i  of  his  capital,  while  B,  having  gained  a  sum 
equal  to  |  of  his  capital,  finds  that  his  money  is  just  equal 
to  that  of  his  brother ;  what  sum  was  given  to  each  ? 

Ans.  A  $1500,  B  $900. 

17.  A  said  to  B,  "give  me  $100,  and  then  I  shall  have 
as  much  as  you."  B  said  to  A,  "give  me  $100,  and 
then  I  shall  have  twice  as  much  as  you."  How  much  had 
each?  Ans.  A  $500,  B  $700. 

18.  If  the  greater  of  two  numbers  be  multiplied  by  5, 
and  the  less  by  7,  the  sum  of  their  products  is  198;  but 
if  the  greater  be  divided  by  5,  and  the  less  by  7,  the  sum 
of  their  quotients  is  6  ;  what  are  the  numbers  ? 

Ans.  20  and  14. 


SIMPLE  EQUATIONS.  123 

19.  Seven  years  ago  the  age  of  A  was  just  three  times 
that  of  B ;  and  seven  years  hence,  A's  age  will  be  just 
double  B's;  what  are  their  ages? 

Ans.  A's  49,  B's  21  yrs. 

20.  There  is  a  certain  number  consisting  of  two  places 
of  figures,  which  being  divided  by  the  sum  of  its  digits, 
the  quotient  is  4,  and  if  27  be  added  to  it,  the  digits  will 
be  inverted ;  required  the  number.  Ans.  36. 

21.  A  grocer  has  two  kinds  of  sugar,  of  such  quality 
that  1  lb.  of  each  are  together  worth  20  cents  ;  but  if  3  lb. 
of  the  first,  and  5  lb.  of  the  second  kind  be  mixed,  a  lb.  of 
the  mixture  will  be  worth  11  cents;  what  is  the  value  of 
a  lb.  of  each  sort?  Ans.  6  cts.,  and  14  cts. 

22.  A  boy  lays  out  84  cents  for  lemons  and  oranges, 
giving  3  cents  apiece  for  the  lemons,  and  5  cents  apiece 
for  the  oranges  ;  he  afterward  sold  i  of  the  lemons  and  ^ 
of  the  oranges  for  40  cents,  and  cleared  8  cents  on  what 
he  sold ;  what  number  of  each  did  he  purchase? 

Ans.  8  lemons,  12  oranges. 

23.  A  owes  $500  and  B  $600,  but  neither  has  suffi- 
cient money  to  pay  his  debts.  A  said  to  B,  "  lend  me 
i  of  your  money,  and  I  can  pay  my  debts."  B  said  to  A, 
"lend  me  \  of  your  money,  and  I  can  pay  mine."  How 
much  has  each?  Ans.  A  $400,  B  $500. 

24.  A  son  said  to  his  father,  "  how  old  are  we?"  The 
father  replied,  "  six  years  ago  my  age  was  3±  times  yours, 
but  3  years  hence  my  age  will  be  only  2i  times  yours." 
Required  their  ages.  Ans.  Father's  36,  son's  15  yrs. 

25.  A  farmer  having  mixed  a  certain  number  of  bu.  of 
oats  and  rye,  found,  that  if  he  had  mixed  6  bu.  more  of 
each,  he  would  have  mixed  7  bu.  of  oats  for  every  6  of 
rye ;  but  if  he  had  mixed  6  bu.  less  of  each,  he  would 
have  put  in  6  bu.  of  oats  for  every  5  of  rye.  How  many 
bu.  of  each  did  he  mix  ?  Ans.  Oats  78,  rye  66  bu. 


124  RAY'S  ALGEBRA,  FIRST  BOOK. 

26.  A  person  having  laid  out  a  rectangular  yard,  ob- 
served that  if  each  side  had  been  4  yd.  longer,  the  length 
would  have  been  to  the  breadth  as  5  to  4 ;  but,  if  each 
had  been  4  yd.  shorter,  the  length  would  have  been  to  the 
breadth  as  4  to  3  ;  required  the  length  of  the  sides. 

Ans.  Length  36,  breadth  28  yd. 

27.  A  farmer  rents  a  farm  for  $245  per  annum ;  the 
tillable  land  being  rented  at  $2  an  acre,  and  the  pasture 
at  $1  and  40  cts.  an  acre  ;  now  the  number  of  acres  till- 
able is  to  the  excess  of  the  tillable  above  the  pasture,  as 
14  to  9 ;  how  many  were  there  of  each  ? 

Ans.  Tillable  98,  pasture  35  A. 

28.  After  drawing  15  gal.  from  each  of  2  casks  of  wine, 
the  quantity  remaining  in  the  first  is  §  of  that  in  the 
second  ;  after  drawing  25  gal.  more  from  each,  the  quan- 
tity left  in  the  first  is  only  half  that  in  the  second ;  what 
number  of  gal.  in  each  before  the  first  drawing? 

Ans.  65  and  90  gal. 

29.  If  1  be  added  to  the  numerator  of  a  certain  frac- 
tion, and  the  numerator  to  the  denominator,  its  value  will 
be  \  ;  but  if  the  denominator  be  increased  by  unity,  and 
the  numerator  by  the  denominator,  its  value  will  be  f  ; 
find  it.  Ans.  T33. 

30.  Find  two  numbers  in  the  ratio  of  5  to  7,  to  which 
two  other  required  numbers,  in  the  ratio  of  3  to  5,  being 
respectively  added,  the  sums  shall  be  in  the  ratio  of  9  to 
13,  and  the  difference  of  their  sums  16. 

Ans.  30  and  42,  6  and  10. 

31.  A  farmer,  with  28  bushels  of  barley,  worth  28  cents 
per  bushel,  would  mix  rye  at  36  cents,  and  wheat  at 
48  cents  per  bushel,  so  that  the  whole  mixture  may  consist 
of  100  bushels,  and  be  worth  40  cents  a  bushel  ;  how  much 
rye  and  wheat  must  be  mixed  with  the  barley  ? 

Ans.  Rye  20,  wheat  52  bushels. 


SIMPLE  EQUATIONS.  125 

32.  A  person  has  two  horses,  and  two  saddles,  one  of 
which  cost  $50,  and  the  other  $2.  If  he  places  the  best 
saddle  upon  the  first  horse,  and  the  other  on  the  second, 
then  the  latter  is  worth  $8  less  than  the  former  ;  but  if 
he  puts  the  worst  saddle  upon  the  first,  and  the  best  upon 
the  second  horse,  then  the  value  of  the  latter  is  to  that 
of  the  former  as  15  to  4  ;  required  the  value  of  each 
horse.  Ans.  First,  $30,  second  $70. 

33.  The  weights  of  two  loaded  wagons  were  in  the  ratio 
of  4  to  5  ;  parts  of  their  loads,  which  were  in  the  ratio 
of  6  to  7,  being  taken  out,  their  weights  were  in  the  ratio 
of  2  to  3,  and  the  sum  of  their  weights  was  then  10  tons  ; 
what  their  weights  at  first?  Ans.  16  and  20  tons. 

34.  A  person  had  two  casks  and  a  certain  quantity  of 
wine  in  each  ;  in  order  to  have  the  same  quantity  in  each 
cask,  he  poured  as  much  out  of  the  first  cask  into  the  second 
as  it  already  contained  ;  he  next  poured  as  much  out  of  the 
second  into  the  first  as  it  then  contained ;  and,  lastly,  he 
poured  out  as  much  from  the  first  into  the  second  as  there 
was  remaining  in  it ;  after  this,  he  had  16  gal.  in  each  cask  ; 
how  many  gal.  in  each  at  first? 

Ans.  First  22,  second  10  gal. 

GENERAL    REVIEW. 

Define  elimination.  How  many  methods  of  elimination?  Of  what 
does  elimination  by  substitution  consist?  Rule.  Elimination  by 
comparison?  Rule.  Elimination  by  addition  and  subtraction? 
Rule.  Define  transposition.  How  are  the  signs  affected  by  trans- 
position?    Explain  by  an  example. 

What  is  the  dimension  of  a  term?  When  is  a  polynomial  homo- 
geneous? For  what  is  a  parenthesis  used?  A  vinculum?  What 
does  the  same  letter  accented  denote?  What  are  similar  or  like 
quantities? 

What  the  rule  for  addition  of  algebraic  quantities?  Subtraction? 
Multiplication?  Division?  Rule  for  the  signs?  For  finding  the 
L.C.M.?     The  G.C.D.?     Define  a  fraction. 

What  the  effect  of  multiplying  the  numerator  of  a  fraction?  The 
denominator?  Both?  Of  dividing  the  numerator?  The  denom- 
inator?    Both?     Repeat  the  axioms. 


126  RAY'S  ALGEBRA,  FIRST  BOOK. 

SIMPLE    EQUATIONS, 

CONTAINING    THREE   OR   MORE   UNKNOWN    QUANTITIES. 

162.  Equations  involving  three  or  more  unknown  quan- 
tities may  be  solved  by  either  of  the  three  methods  of 
elimination  already  explained. 

Suppose  we  have  the  following  equations,  in  which  it  is 
required  to  find  the  values  of  x,  y,  and  z. 

x+2y+  z=20  (1.) 
2x+  !+%*=&!  (2.) 
Sx+4y+2z=U     (3.) 

SOLUTION    BY    SUBSTITUTION. 

From  equation  (1),  a;=20— 2y— z. 
Substituting  this  in  equation  (2),  we  have 

2(20—22/—  z)-fs/+3z=31; 
or,     40— 4y— 2z  -fy+ 3z=31 
3y-z=9     (4.) 

Substituting  the  same  value  of  x  in  equation  (3),  we  have 

3(20— 2y—  z)+4y+2z=44; 
orj    60— 6y— 3z  +4?/+2z=44. 
2t/+z=16     (5.) 
Sij-z=9      (4.) 

The  values  of  y  and  z  are  found,  by  Rule,  Art.  158,  to  be  5  and  6; 
substituting  these  values  in  equation  (1),  X=4. 

SOLUTION   BY   COMPARISON. 


From  equation  (1),  x=20—2y—z. 
Sl—y—Sz 
2         ' 
44— 4y-2z 


(2),  XJ±^. 


(3),  x-. 


SIMPLE  EQUATIONS.  127 

Comparing  the  first  and  second  values  of  X,  we  have 

31—y—Sz 
20-22/-  0= 1 ; 

or,    40— 4y— 2^=31-2/— 30; 
or,    32/-  0=  9     (4.) 

Comparing  the  first  and  third  values  of  x,  we  have 

orj    60— 62/  -30=44— 42/— 20. 
22/+  *=16     (5.) 

From  equations  (4)  and  (5),  the  values  of  y  and  0,  and  then  x, 
may  be  found  by  the  Rule,  Art.  159. 

SOLUTION    BY    ADDITION    AND    SUBTRACTION. 

Multiplying  equation  (1)  by  2,  we  have 

2z-f42/+20=4O 
Equation  (2)  is  2x+  y+3z= 31 
By  subtracting,  3y—  0=  9     (4.) 

Next,  multiplying  equation  (1)  by  3,  we  have 

3x+62/+30=6O 
Equation  (3)  is  3z+42/+20=44 


By  subtracting, 

By  adding, 

22/+  0= 
Sy-  0= 

52/    = 

y   = 

=16 

=  9 

25 
5 

(5.) 
(4.) 

Then, 
And 

10+0=16,  and 
#+10+6=20,  and 

0=6. 
z=4. 

It  will  be  found,  in  practice,  that  the  method  of  elimination  by 
addition  and  subtraction  is  generally  to  be  preferred;  we  shall, 
therefore,  illustrate  it  by  another  example. 

v+2x-{-Sy+4:Z=S0      (1.) 

2*+3a;+  3/+  *=15  (2.) 
3H-  x+2y-\-3z=:23  (3.) 
4u+2»— y+14a=61     (4.) 


128  RAY'S  ALGEBRA,  FIRST  BOOK. 

Let  us  first  eliminate  V;  this  may  be  done  thus : 

2u-j-4flH-  6#-f-  8z=  60,  by  multiplying  equation  (1)  by  2. 
2v-\-3xj-    y-\-    z=  15        (2.) 

*+■  %/+"  72=  45         (5.),  ^  subtracting. 

3u-f-6£-f-  9z/-}-122:=  90,  by  multiplying  equation  (1)  by  3. 
3v-j-  «-j-  3H"  32=  23         (3-) 

5^_|_  7^_|_  qz.-  67          (6.),  by  subtracting. 

4v-j-&r-fl3y-t-163=120,  by  multiplying  equation  (1)  by  4. 
4v-f-2a;—    ff-fl4g=  61         (4.) 

6^4-13//-j-  2s;=  59         (7.),  by  subtracting. 

Collecting  into  one  place  the  new  equations  (5),  (6),  and  (7),  we 
find  that  the  number  of  unknown  quantities,  as  well  as  the  number 
of  equations,  is  one  less. 

xJT  5y-\-7z=45  (5.) 
&z-f-  7?/-}-9z=z67  (6.) 
6x-\-13y-\-2z=59    (7.) 

The  next  step  is  to  eliminate  x,  in  a  similar  manner. 

5#-{-25?/-}-352:=225,  by  multiplying  equation  (5)  by  5. 
5x-\-  ly\-  9z=z  67,  equation  (6). 

18#4-26z=158     (8.),  by  subtracting. 

6x^-30y-\-42z=270,  by  multiplying  equation  (5)  by  6. 
6x-\-13y-\-  2z=  59,  equation  (7). 

17y-^-i0z=211     (9.),  by  subtracting. 

Bringing  together  equations  (8)  and  (9),  the  number  of  equations, 
as  well  as  of  unknown  quantities,  is  two  less. 

18y-ir26z=158    (8.) 
17y-\-40z=211     (9.) 

306?/-|-720z=3798,  by  multiplying  equation  (9)  by  18. 
306>f-442z=2686,  by  multiplying  equation  (8)  by  17. 


27<< 


:1112,  by  subtracting. 


z=      4 

Substituting  the  value  of  z,  in  equation  (9),  we  get 
1 7y-|-160=21 1 ;  and  17^=51 :  and  y=3. 


SIMPLE  EQUATIONS.  129 

Substituting  the  values  of  y  and  z,  in  equation  (5),  we  get 

a4-15-f  28=45 

x=  2 

Substituting  the  values  of  x,  y,  and  Z,  in  equation  (1),  we  have 

V-|_4+9+16=:30 
v=  1 

From  the  preceding  example,  we  derive  the  following 


GENERAL  RULE, 
FOR   ELIMINATION   BY  ADDITION   AND    SUBTRACTION. 

1.  Eliminate  the  same  unknown  quantity  from  each  of  the 
equations;  the  number  of  equations  and  of  unknown  quan- 
tities will  be  one  less. 

2.  Proceed  in  the  same  way  with  another  unknown  quan- 
tity; the  number  of  equations  and  of  unknown  quantities  will 
be  two  less. 

3.  Continue  this  series  of  operations  until  a  single  equa- 
tion is  obtained,  containing  but  one  unknown  quantity. 

4.  By  going  back  and  substituting,  the  values  of  the  other 
unknown  quantities  may  be  readily  found. 

Remark. — When  one  or  more  of  the  equations  contains  but  one 
or  two  of  the  unknown  quantities,  the  method  of  substitution  will 
generally  be  found  the  shortest. 

In  literal  equations  and  some  others,  the  method  of  comparison 
may  be  most  convenient.  After  solving  several  examples  by  each 
method,  the  pupil  will  be  able  to  appreciate  their  relative  excellence 
in  different  cases. 


SOLVE   BY    EITHER   METHOD    OF    ELIMINATION. 

x+y=$0 rx=lS. 

*+z=28 Ans.  ]y=32. 

y+z=±2 U=10. 


130 


RAY'S  ALGEBRA,  FIRST  BOOK. 


4. 


5. 


6. 


Zx+by=  76 rx=12. 

4z+6z=108 Ans.    |y=    8. 

5*  +  7y=106 U=10. 

x-\-y-\-z=  26 p-    3. 

x-\-y — 8= — 6 Ans.    •}  y=    7. 

x—y+z=  12 (2=16. 


aj+|=100. 
y+|==lQ0. 


*+=f=100. 
4 


<c=64. 

*=84. 


2x—  y-\-  z=  9 rz=3. 

a>— 2y+3«=14 Ans.   \y=2. 

3z-f4y— 2z=  7 M==5. 


___  +  ,_3. 
6^4      S 

x    y  i      k 

2-3  +  *=5- 


Ans.    - 


2:^6. 
#=4. 

Z=S. 


PROBLEMS  PRODUCING  EQUATIONS  CONTAINING  THREE  OR 
MORE  UNKNOWN  QUANTITIES. 

163.  When  a  problem  contains  three  or  more  unknown 
quantities,  the  equations  may  be  formed  according  to  the 
directions  given  in  Art.  156  and  161. 

Remark. — When  one  or  more  of  the  unknown  quantities  can  be 
expressed  in  terms  of  another,  it  is  best  to  reduce  the  number  of 
equations  and  symbols  by  doing  so. 

Review. — 162.  What  is  the  general  n:ie  for  elimination  by  addi- 
tion and  subtraction?  When  is  elimination  by  substitution  to  be 
preferred?     When  that  by  comparison? 


SIMPLE  EQUATIONS.  131 

1.  A  has  3  ingots,  composed  of  different  metals  in  dif- 
ferent proportions  ;  1  lb.  of  the  first  contains  7  oz.  of  sil- 
ver, 3  of  copper,  and  6  of  tin ;  1  lb.  of  the  second  con- 
tains 12  oz.  of  silver,  3  of  copper,  and  1  of  tin  ;  and  1  lb. 
of  the  third  contains  4  oz.  of  silver,  7  of  copper,  and  5  of 
tin.  How  much  of  each  must  be  taken  to  form  an  ingot 
of  1  lb.  weight,  containing  8  oz.  of  silver,  3|  of  copper, 
and  4|  of  tin  ? 

Let  X,  y,  z,  represent  the  number  of  oz.  taken  of  the  3  ingots 

respectively. 

Then,  since  16  oz.  of  the  first  contains  7  oz.  of  silver,  1  oz.  will 

7x 
contain  X  oz.  of  silver;   and  X  oz.  will  contain  ^  oz.  of  silver 

16        12y 

In  like  manner,  y  oz.  of  the  second  will  contain    -^f-  oz.  of  sil- 

4z  16 

ver;  and  z  oz.  of  the  third  will  contain  «  oz.  of  silver. 

Jo 

But,  by  the  question,  the  number  of  oz.  of  silver  in  a  pound  of 
the  new  ingot,  is  to  be  8;  hence, 

16+  16  "•"IB 

Or,  by  clearing  it  of  fractions, 

7z+12#+4z=128     (1.) 

Reasoning  in  a  similar  manner  with  reference  to  the  copper  and 
the  tin,  we  have  the  two  following  equations : 

3:r-f3?/+7z=60     (2.) 
6x+  ?/+5z=68     (3.) 

The  terms  containing  y  being  the  simplest,  will  be  most  easily 
eliminated. 

Multiplying  (2)  by  4,  and  subtracting  (1),  we  have 

5^4-240=112     (4.) 

Multiplying  (3)  by  3,  and  subtracting  (2),  we  have 

15z+8z=144    (5.) 

Review. — 163.  Upon  what  principle  are  equations  formed,  when  a 
problem  contains  three  or  more  unknown  quantities  ?  When  may 
we  reduce  the  number  of  symbols? 


132  RAY  S  ALGEBRA,  FIRST  BOOK. 

Multiplying  (5)  by  3,  and  subtracting  (4),  there  results 
40x=320 

Substituting  this  value  of  x  in  equation  (5),  we  have 
120+8z=144 
z=3 

And  substituting  these  values  of  x  and  z  in  equation  (3), 
48-f2/_|-15=68 

Hence,  the  new  ingot  will  contain  8  oz.  of  the  first,  5  of  the  second, 
and  3  of  the  third. 


2.  The  sums  of  three  numbers,  taken  two  and  two,  are 
27,  32,  and  35  ;  required  the  numbers. 

Ans.  12,  15,  and  20. 

3.  The  sum  of  three  numbers  is  59 ;  ^  the  difference 
of  the  first  and  second  is  5,  and  ^  the  difference  of  the 
first  and  third  is  9  ;  required  the  numbers. 

Ans.  29,  19,  and  11. 

4.  A  person  bought  three  silver  watches ;  the  price  of 
the  first,  with  J  the  price  of  the  other  two,  was  $25 ;  the 
price  of  the  second,  with  i  the  price  of  the  other  two, 
was  $26 ;  and  the  price  of  the  third,  with  J  the  price  of 
the  other  two,  was  $29 ;  required  the  price  of  each. 

Ans.  $8,  $18,  and  $16. 

5.  Find  three  numbers,  such  that  the  first  with  J  of  the 
other  two,  the  second  with  \  of  the  other  two,  and  the 
third  with  |  of  the  other  two,  shall  each  equal  25. 

Ans.  13,  17,  and  19. 

6.  A  boy  bought  at  one  time  2  apples  and  5  pears, 
for  12  cts. ;  at  another,  3  pears  and  4  peaches,  for  18  cts. ; 
at  another,  4  pears  and  5  oranges,  for  28  cts. ;  and  at  an- 
other, 5  peaches  and  6  oranges,  for  39  cts. ;  required  the 
cost  of  each  kind  of  fruit. 

Ans.  Apples  1,  pears  2,  peaches  3,  oranges  4  cts.  each. 


SIMPLE  EQUATIONS.  133 

7.  A  and  B  together  possess  only  §  as  much  money 
as  C ;  B  and  C  together  have  6  times  as  much  as  A ;  and 
B  has  $680  less  than  A  and  C  together ;  how  much  has 
each?  Ans.  A  $200,  B  $360,  and  C  $840. 

8.  A,  B,  and  C  compare  their  money;  A  says  to  B, 
"give  me  $700,  and  I  shall  have  twice  as  much  as  you 
will  have  left."  B  says  to  C,  "give  me  $1400,  and  I  shall 
have  three  times  as  much  as  you  will  have  left."  C  says 
to  A,  "  give  me  $420,  and  I  shall  have  five  times  as  much 
as  you  will  have  left."     How  much  has  each? 

Ans.  A  $980,  B  $1540,  and  C  $2380. 

9.  A  certain  number  is  expressed  by  three  figures,  whose 
sum  is  11  ;  the  figure  in  the  place  of  units  is  double  that 
in  the  place  of  hundreds  ;  and  if  297  be  added  to  the  num- 
ber, its  figures  will  be  inverted ;  required  the  number. 

Ans.  326. 

10.  The  sum  of  3  numbers  is  83  ;  if  from  the  first  and 
second  you  subtract  7,  the  remainders  are  as  5  to  3 ; 
but  if  from  the  second  and  third  you  subtract  3,  the  re- 
mainders are  to  each  other  as  11  to  9 ;  required  the 
numbers.  Ans.  37,  25,  and  21. 

11.  Divide  $180  among  three  persons,  A,  B,  and  C, 
so  that  twice  A's  share  plus  $80,  three  times  B's  share 
plus  $40,  and  four  times  C's  share  plus  $20,  may  be  all 
equal  to  each  other.  Ans.  A  $70,  B  $60,  C  $50. 

12.  If  A  and  B  can  perform  a  certain  work  in  12  days, 
A  and  C  in  15  days,  and  B  and  C  in  20  days,  in  what 
time  could  each  do  it  alone? 

Ans.  A  20,  B  30,  and  C  60  days. 

13.  A  number  expressed  by  three  figures,  when  divided 
by  the  sum  of  the  figures  plus  9,  gives  a  quotient  of  19  ; 
the  middle  figure  equals  half  the  sum  of  the  first  and 
third ;  and  if  198  be  added  to  the  number,  we  obtain  a 
number  with  the  same  figures  in  an  inverted  order ;  what 
is  the  number?  Ans.  456. 


134  RAY'S  ALGEBRA,  FIRST  BOOK. 

14.  A  farmer  mixes  barley  at  28  cents,  with  rye  at  36, 
and  wheat  at  48  cents  per  bu.,  so  that  the  whole  is  100  bu., 
and  worth  40  cents  per  bu.  Had  he  put  twice  as  much 
rye,  and  10  bu.  more  wheat,  the  whole  would  have  been 
worth  exactly  the  same  per  bu.  ;  how  much  of  each  was 
there?  Ans.  Barley  28,  rye  20,  wheat  52  bu. 

15.  A,  B,  and  C  killed  90  birds,  which  they  wish  to 
share  equally  ;  to  do  this,  A,  who  has  the  most,  gives  to  B 
and  C  as  many  as  they  already  had  ;  next,  B  gives  to  A  and 
C  as  many  as  they  had  after  the  first  division  ;  lastly,  C 
gives  to  A  and  B  as  many  as  they  both  had  after  the  second 
division,  and  each  then  had  the  same  number ;  how  many 
had  each  at  first?  Ans.  A  52,  B  28,  and  C  16. 


GENERAL    REVIEW. 

What  two  parts  in  the  solution  of  a  problem?  What  are  explicit 
conditions?  Implied  conditions?  Rule  for  forming  an  equation. 
On  what  condition  may  you  change  the  sign  of  one  term  in  an 
equation? 

Define  elimination.  How  many  methods  of  elimination?  Define 
elimination  by  substitution — by  comparison — by  addition  and  sub- 
traction. Rule  for  each  method.  How  state  a  problem  containing 
two  unknown  quantities?  How  one  containing  three  or  more  un- 
known quantities?  When  is  the  first  method  of  elimination  pre- 
ferred? When  the  second?  The  third?  Rule  for  elimination  in 
three  or  more  unknown  quantities. 

Give  two  rules  for  rendering  a  complex  fraction  simple.  State 
the  eight  theorems,  Arts.  80  to  85.  Rule  for  exponents  in  multipli- 
cation. In  division.  Difference  between  subtraction  in  algebra  and 
in  arithmetic.  In  clearing  an  equation  of  fractions,  what  is  to  be 
done  when  there  is  a  minus  sign  before  a  fraction? 

Define  binomial.  Term.  Coefficient.  Exponent.  Factor.  Prime 
number.  Composite  number.  What  is  the  reciprocal  of  a  fraction? 
What  are  the  factors  of  x2—l,  of  a;3— 1,  of  #3-f-l,  of  x2+3x-\- 2? 
By  how  many  different  methods  could  you  reduce  ^,  l,  |,  and  -3L  to 
a  common  denominator? 

In  >vhat  cases  may  cancellation  be  employed  to  advantage? 
What  three  methods  of  multiplying  a  fraction  by  a  whole  number? 
Of  dividing  a  fraction  by  a  whole  number?  What  are  infinite 
series  ?  What  the  law  of  a  series  ?  How  convert  -^y  into  an  infinite 
series? 


GENERALIZATION.  135 

V.     SUPPLEMENT     TO     SIMPLE 
EQUATIONS. 

GENERALIZATION. 

164.  A  Literal  Equation  is  an  equation  in  which  the 
known  quantities  are  represented,  either  entirely  or  partly, 
by  letters. 

Values  expressed  by  letters  are  termed  general,  because, 
by  giving  particular  values  to  the  letters,  the  solution  of 
one  problem  furnishes  a  general  solution  to  all  others  of 
the  same  kind. 

A  Formula  is  the  answer  to  a  problem,  when  the  known 
quantities  are  represented  by  letters. 

A  Rule  is  a  formula  expressed  in  ordinary  language. 

By  the  application  of  Algebra  to  the  solution  of  general 
questions,  a  great  number  of  useful  and  interesting  truths 
and  rules  may  be  established. 

We  now  proceed  to  illustrate  this  subject  by  a  few  ex- 
amples. 

1G*>. — 1.  Let  it  be  required  to  find  a  number,  which 
being  divided  by  3,  and  by  5,  the  sum  of  the  quotients 

will  be  16. 

cc    cc 
Let  £=  the  number;  then,  ^-(-==16. 
o     o 

5#+3a;=16xlo 

&c=16xl5 

x=  2x15=30 

2.  Again,  let  it  be  required  to  find  another  number, 
which  being  divided  by  4,  and  by  7,  the  sum  of  the  quo- 
tients will  be  11. 

By  proceeding  as  in  the  above  question,  we  find  the  number 
to  be  28. 


136  RAYS  ALGEBRA,  FIRST  BOOK. 

Instead  of  solving  every  example  of  the  same  kind  separately, 
we  may  give  a  general  solution,  that  will  embrace  all  the  particular 
questions;   thus: 

3.  Let   it  be  required  to  find  a   number,  which  being 

divided  by  two  given  numbers,  a  and  b,  the  sum  of  the 

quotients  may  be  equal  to  another  given  number,  c. 

X     x 
Let  X—  the  number :   then,  — \-T=c. 
'  a  '  b 

bx-\-ax=iabc 

{a-\-b)x=abc 

abc 

The  answer  is  termed  a  formula;  it  shows  that  the  required  num- 
ber is  equal  to  the  continued  product  of  a,  b,  and  C,  divided  by  the 
sum  of  a  and  b.     Or,  it  may  be  expressed  thus : 

Multiply  together  the  three  given  numbers,  and  divide  the 
product  by  the  sum  of  the  divisors;  the  result  will  be  the  re- 
quired number. 

The  pupil  may  test  the  accuracy  of  this  rule  by  solving 
the  following  examples,  and  verifying  the  results : 

4.  Find  a  number  which  being  divided  by  3,  and  by  7, 
the  sum  of  the  quotients  may  be  20.  Ans.  42. 

5.  Find  a  number  which  being  divided  by  |  and  J,  the 
sum  of  the  quotients  may  be  1 .  Ans.  J,. 

166.— 1.  The  sum  of  $500  is  to  be  divided  between 
two  persons,  A  and  B,  so  that  A  may  have  $50  less 
than  B.  Ans.  A  $225,  B  $275. 

To  make  this  question  general,  let  it  be  stated  as  follows  : 

Review. — 164.  What  is  a  literal  equation?  When  are  values 
termed  general?  What  is  a  formula?  What  is  a  formula  called 
when  expressed  in  ordinary  language? 

165.  Example  3.  What  is  the  answer  to  this  question,  expressed 
in  ordinary  language? 


GENERALIZATION.  137 

2.  To  divide  a  given  number,  a,  into  two  such  parts, 
that  their  difference  shall  be  b  ;  or,  the  sum  of  two  num- 
bers is  a,  and  their  difference  h ;  required  the  numbers. 

Let  £=  the  greater  number,  and  y~  the  less. 

Then,  x+y=a 

And  x—y~b 
By  addition,    2x=a+b 

a-f-6 a    b 

By  subtraction,  2y=a — b 

a—b    a     b 

This  formula,  expressed  in  ordinary  language,  gives  the  following 


RULE, 

FOR    FINDING    TWO     QUANTITIES,    WHEN    THEIR    SUM     AND 
DIFFERENCE    ARE    GIVEN. 

1.  To  find  the  greater,   add   half  the  difference  to  half 
the  sum. 

2.  To  find  the  less,  subtract  half  the  difference  from  half 
the  sum. 

Test  the  accuracy  of  the  rule,  by  finding  the  two  num- 
bers in  the  following  examples : 

3.  Sum  200,  difference  50 Ans.  125,     75. 

4.  Sum  100,  difference  25 Ans.  62A,  37J. 

5.  Sum     15,  difference  10 Ans.  12j,     2j. 

6.  Sum     5£,  difference     f Ans.     3J,     2j. 

167. — 1.  A  can  do  a  piece  of  work  in  3  da.,  and  B 
in  4  da. ;  in  what  time  can  both  together  do  it  ? 

Ans.  If  da. 

To  make  this  question  general,  let  it  be  stated  thus : 
1st  Bk.  12 


138  RAY'S  ALGEBRA,  FIRST  BOOK. 

2.  A  can  do  a  piece  of  work  in  m  da.,  and  B  in  n  da.;  in 
how  many  da.  can  they  both  together  do  it  ? 

Let  x=  the  number  of  da.  in  which  they  can  both  do  it. 

Then,   -=  the  part  of  the  work  which  both  can  do  in  one  da. 

X  1  1 

Also,  A  can  do  —  part  and  B  can  do  -  part  of  it  in  1  da.     Hence, 

the  part  of  the  work  which  both  can  do  in  1  da.  is  represented  by 

1 — ,  and  also  bv  -. 

m    n  "  x  111 

Therefore, _ _i_^=_. 

m  '  n    x 

nx-±-mx=mn 

mn 

X=z- 


ra-j-n 
This  result,  expressed  in  ordinary  language,  gives  the  following 

Rule. — Divide  the  product  of  the  numbers  expressing  the 
time  in  which  each  can  perform  the  work,  by  their  sum;  the 
quotient  will  be  the  time  in  which  they  can  jointly  perform  it. 

The  question  can  be  made  more  general,  thus : 

A  can  produce  a  certain  effect,  e,  in  a  time,  t ;  B  can  pro- 
duce the  same  effect,  in  a  time,  t'\  in  what  time  can  they 
both  do  it? 

The  result  and  the  rule  would  be  the  same  as  already 
given. 

The  following  examples  will  illustrate  the  rule : 

3.  A  cistern  is  filled  by  one  pipe  in  6,  and  by  another 
in  9  hr. ;  in  what  time  will  it  be  filled  by  both  together  ? 

Ans.  3^  hr. 

o 

4.  One  man  can  drink  a  keg  of  cider  in  5  da.,  and  an- 
other in  7  da. ;  in  what  time  can  both  together  drink  it  ? 

Ans.  2-U  da. 

Review. — 166.  By  what  rule  do  we  find  two  quantities,  when  their 
sura  and  difference  are  given? 

167.  When  the  times  are  given,  in  which  each  of  two  men  can 
produce  a  certain  effect,  how  is  the  time  found  in  which  they  can 
jointly  produce  it? 


GENERALIZATION.  139 

1G8.  Let  it  be  required  to  find  a  rule  for  dividing  the 
gain  or  loss  in  a  partnership.  First,  take  a  particular 
question. 

1.  A,  B,  and  C  engage  in  trade,  and  put  in  stock  in  the 
following  proportions  :  A  put  in  $3  as  often  as  B  put 
in  $4,  and  as  often  as  C  put  in  $5.  They  gain  $60  ;  re- 
quired the  share  of  each,  it  being  divided  in  proportion  to 
the  stock  put  in. 

Let  Sx=  A's  share  of  the  gain;  then,  4#=  B's,  and  5x=  C's. 
(See  Example  24,  page  126.) 
Then,  3x+4x+ 5z=60. 
12z=60 
x=  5 
3#=15,  A's  share;  4#=20,  B's;  and  5#=25,  C's  share. 

2.  To  make  this  question  general,  suppose  A  puts  in 
m  $'s  as  often  as  B  puts  in  n  $'s,  and  as  often  as  C  puts 
in  r  $'s,  and  that  they  gain  c  $'s.     Eind  the  share  of  each. 

Let  the  share  of  A  be  denoted  by  mx;  then,  nx==  B's,  and  rx=  C's 
share.    Then,  mx-\-nx-\-rx—C. 

mc  nc  re 

rx=- 


m^-n-^r'  m-\-n-\-r'  m-\-n-\-r 

If  c  had  represented  loss  instead  of  gain,  the  same  solution  would 
have  applied.  Hence,  to  find  each  partner's  share  of  the  gain  or 
loss,  we  have  the  following 

Rule. — Divide  (he  whole  gain  or  loss  by  the  sum  of  the 
proportions  of  stock,  and  multiply  the  quotient  by  each  part- 
ner's proportion,  to  obtain  his  respective  share. 

When  the  times  in  which  the  respective  stocks  are  em- 
ployed are  different,  it  becomes  necessary  to  reduce  them 
to  the  same  time,  to  ascertain  what  proportion  they  bear 
to  each  other. 

Review. — 168.  How  is  the  gain  or  loss  in  fellowship  found,  when 
the  times  in  which  the  stock  is  employed  are  the  same?  How,  when 
different  ? 


140  RAY'S  ALGEBRA,  FIRST  BOOK. 

Thus,  if  A  have  $3  in  trade  4  mon.,  and  B  $2  for  5  mon., 
we  see  that  $3  for  4  mon.  are  the  same  as  $12  for  1  mon.; 
and  $2  for  5  mon.  are  the  same  as  $10  for  1  mon. 
Therefore,  in  this  case,  the  stocks  are  in  the  proportion 
of  12  to  10. 

Hence,  when  time  in  fellowship  is  considered,  we  have 
the  following 


Rule. — Multiply  each  man's  stock  by  the  time  it  was  em- 
ployed to  find  the  proportions  of  stock;  and  then  proceed 
according  to  the  preceding  rule. 

3.  A,  B,  and  C  engaged  in  trade ;  A  put  in  $200, 
B  $300,  and  C  $700;  they  lost  $60  ;  what  was  each  man's 
share?  Ans.  A's  loss  $10,  B's  $15,  C's  $35. 

Since  the  sums  engaged  are  to  each  other  as  2,  3,  and  7,  we  may 
either  use  these  numbers,  or  those  representing  the  stock. 

4.  In  a  trading  expedition,  A  put  in  $200  for  3  mon., 
B  $150  for  5  mon.,  and  C  $100  for  8  mon.;  they  gained 
$215  ;  what  was  each  man's  share? 

Ans.  A's  $60,  B's  $75,  C's  $80. 

169. — 1.  Two  men,  A  and  B,  can  perform  a  certain 
piece  of  work  in  a  da.,  A  and  C  in  b  da.,  and  B  and  C 
in  c  da.;  in  what  time  could  each  one  alone  perform  it? 
In  what  time  could  they  perform  it,  all  working  to- 
gether ? 

Let  X,  y,  and  z  represent  the  days  in  which  A,  B,  and  C  can 
respectively  do  it. 

Then,  -,  -,  and  -,  represent  the  parts  of  the  work  which  A,  B, 

and  C  can  each  do  in  1  da. 

Since  A  and  B  can  do  it  in  a  da.,  they  do  —  part  of  it  in  1  da. 


GENERALIZATION.  141 


Hence,    — t— =—     (1).     In  like  manner,  we  have 
'   x^y    a     v  ' 

— I \— =o — h«r+s-     (4),  dividing  by -. 

x^y^z    2a^2b^2c     \.*  °    ^ 

11,1        1     6c-}-ac— a&       ~  M. 

x-25+26-2-c=       2abc      '  ™btractln*  (3)  from  M  I 
or,  £(ac-{-6c — ab)=2abc,  by  clearing  of  fractions. 
2a6c 


t     ,.,  2a6c 

In  like  manner,  y=    ,  ,  . . 

'  a     ab-\-oe — ac 

2abc 

And  z=—r-l r-. 

«6-{-ac — be 

Since  — | — •) — ,  or  ■= — h"5x4"s^i  represents  the  part  all  can  do  in 

,.  -,    1  i.    /   *   .    !   ,  1   \    i           •              2a6c 
1  da. ;  if  we  divide  1  by  I  s — r?n:-rs-   I)  the  quotient,  -*—. r-r- , 

J  \  2a  l  2b  *  2c  r  '  ab-{-ac-\-bc 

will  represent  the  number  of  da.  in  which  all  can  do  it. 

170.  In-  solving  questions,  it  is  sometimes  necessary 
to  use  general  values  for  particular  quantities,  to  ascertain 
the  relation  which  they  bear  to  each  other  ;  as  in  the  fol- 
lowing : 

If  4  A.  pasture  40  sheep  4  wk.,  and  8  A.  pasture  56 
sheep  10  wk.,  how  many  sheep  will  20  A.  pasture  50  wk., 
the  grass  growing  uniformly  all  the  time? 

The  chief  difficulty  in  solving  this  question,  consists  in  ascer- 
taining the  relation  that  exists  between  the  original  quantity  of  grass 
on  an  A.,  and  the  growth  on  each  A.  in  1  wk. 

Let  m=  the  quantity  on  an  A.  when  the  pasturage  began,  and  n= 
the  growth  on  1  A.  in  1  wk. ;  m  and  n  representing  lb.;  or  any  other 
measure  of  the  quantity  of  grass. 


142  RAY'S  ALGEBRA,  FIRST  BOOK. 

Then,  4ra=  the  growth  on  1  A.  in  4  wk. 

And  16ra=  the  growth  on  4  A.  in  4  wk. 

Also,  4ra-j-16n=  the  whole  amount  of  grass  on  4  A.  in  4  wk. 

If  40  sheep  eat  4ra-f-16ra  in  4  wk.,  then  40  sheep  will  eat 

4ra4-16n         ,  . 

L =w_j_4n  in  i  wk 

m-\-4n_  m     n 
~W~~  40+10 
Again,  8ra-{-80n=  the  whole  amount  of  grass  on  8  A.  in  10  wk. 

If  56  sheep  eat  8ra-{-80ra  in  10  wk., 

Then,  56  sheep  eat  ^  -\-Sn  in  1  wk. 

.     ,  1    ,  8ra.8ram.ra..,, 

And  1  sheep  eats  m^=^-f  »  1  wk. 

TT  ra  .  ra     ra  .  ra 

Hence'  40+T0=70+7- 
Or,  7w-f  28n=4ra-|-40ra. 
3m=12ra 
m=4ra 

Or,   ra=}ra;    hence,    the   growth   on   1    A.   in   1    wk.,   is   equal 
to  \  of  the  original  quantity  on  1  A. 
_,        -     ,         .     ,  ra  ,  ra     ra  .  m    in 

Then,  1  sheep,  in  1  wk.,  eats  4o+Io=40+40=20' 

ra  5ra 

And  1  sheep,  in  50  wk.,  eats,  ~\b0—— 

20  A.  have  an  original  quantity  of  grass,  denoted  by  20ra. 

50ra 
The  growth  of  1  A,  in  1  wk.  being  ^ra,  in  50  wk.  it  will  be  — .— 

50ra  4 

And  the  growth  of  20  A  in  50  wk.,  will  be  -j-X20=250ra. 

Then,  20ra-f250ra=270ra,  the  whole  amount  of  grass  on  20  A. 

in  50  wk. 

5ra    540ra 
Then,  210m-. — 0-=—- — -=108,  the  number  of  sheep  required. 
A       o?ra 


GENERAL     PROBLEMS. 

1.  Divide  the  number  a  into  two  parts,  so  that  one  of 

them  shall  be  n  times  the  other.  na  .      a 

Ans.  — r-^-  and  — — T. 
ra+1  ra+l 


GENERALIZATION.  143 

2.  Divide  the  number  a  into  two  parts,  so  that  m  times 
one  part  shall  be  equal  to  n  times  the  other. 

na  .     ma 

Ans.  - — —  and 


m-\-n  m-\-n 

3.  Find  a  number  which  being  divided  by  m,  and  by  n, 
the  sum  of  the  quotients  shall  be  equal  to  a.  mna 

'  m~\-n 

4.  What  number  must  be  added  to  a  and  ?>,  so  that  the 

sums  shall  be  to  each  other  as  m  to  n  ?  .        wh — na 

Ans. . 

u — m 

5.  What  number  must  be  subtracted  from  a  and  6,  so 
that  the  differences  shall  be  to  each  other  as  m  to  w? 

:        na — mb 

Ans. . 

n — m 

6.  After  paying  away  —  and  -  of  my  money,  I  had  a 

dollars  left  ;  how  many  dollars  had  I  at  first  ? 

mva 

Ans.   ■ 

mn — m — n 

7.  A  company  paid  for  the  use  of  a  boat  for  an  excur- 
sion, a  cents  each  ;  if  there  had  been  b  persons  less,  each 
would  have  had  to  pay  c  cents ;  how  many  persons  were 

there  ?  be 

Ans. . 


o — a 

8.  A  farmer  mixes  oats  at  a  cents  per  bu.,  with  rye  at 

b  cents  per  bu.,  so  that  a  bu.  of  the  mixture  is  worth  c 

cents ;  how  many  bu.  of  each  will   n  bu.  of  the  mixture 

contain?  '        n(c — b)       ,  n(a—c) 

Ans.  -^ ^  and  -^ — r-y. 

a — 6  a — 6 

9.  A  person  borrowed  as  much  money  as  he  had  in  his 

purse,  and  then  spent  a  cents ;  again,  he  borrowed  as  much 

as  he  had  in  his  purse,  after  which  he  spent  a  cents  ;  he 

borrowed  and  spent,  in  the  same  manner,  a  third  and  fourth 

time,  after  which,  he  had  nothing  left  j  how  much  had  he 

at  first?  15a 

Ans.Tr 


144  RAYS  ALGEBRA,  FIRST  BOOK. 

10.  A  person  has  2  kinds  of  coin :  it  takes  a  pieces  of 
the  first,  and  b  pieces  of  the  second,  to  make  $1 ;  how- 
many  pieces  of  each  kind  must  be  taken,  so  that  c  pieces 

may  be  equivalent  to  $1  ?  a(b — c)       ,  b(c — a) 

Ans.  -^ and  — ^ \ 

0 — a  0 — a 

1T1.  Sometimes  in  an  equation  of  the  first  degree,  the 
second,  or  some  higher  power  of  the  unknown  quantity 
occurs,  but  in  such  a  manner  that  it  may  be  made  to  dis- 
appear.     The  following  examples  belong  to  this  class  : 

1.  Given  2x2-{-Sx=llx2 — 10x,  to  find  the  value  of  x. 

By  dividing  each  side  by  x,  we  have 

2^+8=11^—10,  from  which  z=2. 

2.  Given  (4+a0(3+a:)— 6(10— x)=x(1+x),  to  find  x. 
Performing  the  operations  indicated,  we  have 

12-f7z+z2— 60-f6a;=7z-fa;2. 

Omitting  the  quantities  on  each  side  which  are  equal,  we  have 

12— 60+6a;=0,  from  which  x=8. 

3.  Sx2—Sx=24x—bx2 Ans.  x=4. 

4.  Sax8—  I0ax2=8ax2+ax*. .    ■ Ans.  x=9. 

i    6x4-13       3.T+5      2x  .  ork 

5.  — =-= = n-=— Ans.  x=Z\). 

10  hx—  25       5 

6.  (a+x)(b+x)— a(c— b)=x(b+x).       Ans.  x=c—2b. 

H        ,       ,  ,   :         x2-f-a2-}-Z>2+c2  ,  c2— ab 

7.  x-\-a-{-b-\-c= p^ = .      .     .     Ans.  x= — r-r- . 

a-f-6 — c-j-x  a-yb 

8.  If  a  certain  book  had  5  more  pages,  with  10  more 
lines  on  a  page,  the  number  of  lines  would  be  increased 
450  ;  if  it  had  10  pages  less,  with  5  lines  less  on  a  page, 
the  whole  number  of  lines  would  be  diminished  450  ;  re- 
quired the  number  of  pages,  and  of  lines  on  a  page. 

Ans.  20  pages,  40  lines  on  a  page. 


NEGATIVE  SOLUTIONS.  145 

NEGATIVE     SOLUTIONS. 

172.  It  sometimes  happens,  in  the  solution  of  a  prob- 
lem, that  the  value  of  the  unknown  quantity  is  found  to 
he  minus.  Such  a  result  is  termed  a  negative  solution.  We 
shall  now  examine  a  question  of  this  kind. 

1.  What  number  must  be  added  to  the  number  5,  that 
the  sum  shall  be  equal  to  3  ? 

Let  x=  the  number. 

Then,  5-f  x=S;  and  X—S— 5==— 2. 

Now,  —2  added  to  5,  gives  3;  thus,  5-f- (—2)  =3.  The  result, 
— 2,  is  said  to  satisfy  the  question  in  an  algebraic  sense;  but  the 
problem  is  evidently  impossible  in  an  arithmetical  sense. 

Since  adding  — 2  is  the  same  as  subtracting  -f  2,  Art.  61,  the  re- 
sult is  the  answer  to  the  following  question:  What  number  must 
be  subtracted  from  5,  that  the  remainder  may  be  equal  to  3  ? 

Let  the  question  now  be  made  general,  thus : 

What  number  must  be  added  to  the  number  a,  that  the 
sum  shall  be  equal  to  b  ? 

Let  x—  the  number.     Then,  a+x=b  ;  and  x—b—a. 

Now,  since  a-f  (6 — Gf)=6,  this  value  of  x  will  always  satisfy  the 
question  in  an  algebraic  sense. 

While  b  is  greater  than  a,  the  value  of  X  will  be  positive,  and  the 
question  will  always  be  consistent  in  an  arithmetical  sense.  Thus, 
if  6=10,  and  a=8,  then  x=2. 

When  b  is  less  than  a,  the  value  of  x  will  be  negative;  the  ques- 
tion will  then  be  true  in  its  algebraic,  but  not  in  its  arithmetical 
sense,  and  should  be  stated  thus :  What  number  must  be  subtracted 
from  a,  that  the  remainder  may  be  equal  to  6?     Hence, 

1.  A  negative  solution  indicates  some  inconsistency  or  ab- 
surdity in  the  question  from  which  the  equation  was  derived. 

2.  When  a  negative  solution  is  obtained,  the  question  to 
which  it  is  the  answer  may  be  so  modified  as  to  be  consistent. 

Keview. — 172.  What  is  a  negative  solution?     When  is  a  result 
said  to  satisfy  a  question  in  an  algebraic  sense?    In  an  arithmetical 
sense?     What  does  a  negative  solution  indicate? 
1st  Bk.  13* 


146  RAY'S  ALGEBRA,  FIRST  BOOK. 

Let  the  pupil  now  read  the  Observations  on  Addition 
and  Subtraction,  page  27,  and  then  modify  the  follow- 
ing questions,  so  that  they  shall  be  consistent  in  an  arith- 
metical sense. 

2.  What  number  must  be  subtracted  from  20,  that  the 
remainder  shall  be  25  ?     (x= — 5.) 

3.  What  number  must  be  added  to  11,  that  the  sum  being 
multiplied  by  5,  the  product  shall  be  40  ?     (ax= — 3.) 

4.  What  number  is  that  whose  |  exceeds  its  {  by  3  ? 

(x=— 36.) 
5-  A  father,  whose  age  is  45  yr.,  has  a  son  aged  15  ;  in 
how  many  yr.  will  the  son  be  \  as  old  as  his  father? 

DISCUSSION    OF    PROBLEMS. 

173.  When  a  question  has  been  solved  in  a  general 
manner,  that  is,  by  representing  the  known  quantities  by 
letters,  we  may  inquire  what  values  the  results  will  have 
when  particular  suppositions  are  made  with  regard  to  the 
known  quantities. 

The  determination  of  these  values,  and  the  examination 
of  the  results,  constitute  what  is  termed  the  discussion  of 
the  problem. 

Let  us  take,  for  example,  the  following  question  : 

1.  After  subtracting  h  from  a,  what  number,  multiplied 
"by  the  remainder,  will  give  a  product  equal  to  c? 

Let  JE=  the  number. 

Then,  (a— b)x=c,   and  x= =-. 

K         '  a—b 

Review. — 172.  When  a  negative  solution  is  obtained,  how  may 
the  question  to  which  it  is  the  answer  be  modified? 

173.  What  is  understood  by  the  discussion  of  a  problem?  The 
expression  c  divided  by  a — b,  may  have  how  many  forms?  Name 
them. 


DISCUSSION  OF  PROBLEMS.  147 

This  result  may  have  five  different  forms,  depending  on 
the  values  of  a,  6,  and  c. 

We  shall  examine  each  of  these  in  succession. 

I.  When  b  is  less  than  a. 

In  this  case,  a — b  is  a  positive  quantity,  and  the  value 
of  x  is  positive. 

To  illustrate  this  form,  let  a-=S,  6=3,  and  C=20 ;  then,  X=4. 

II.  When  b  is  greater  than  a. 

In  this  case,  a — b  is  a  negative  quantity,  and  the  value 
of  x  will  be  negative. 

To  illustrate  this  case  by  numbers,  let  a=2,  6=5,  and  c=12; 
then,  a — 6= — 3,  and  #= — 4. 

III.  When  b  is  equal  to  a. 

In  this  case,  x  becomes  equal  to  ^. 

We  must  now  inquire,  what  is  the  value  of  a  fraction  when  the 
denominator  is  zero. 

If  we  divide  C  successively  by  1,  y\j,  T^j,  yo\nj>  etc.,  the  quo- 
tients will  be  c,  10c,  100c,  1000c,  etc. 

As  the  denominator  decreases,  the  value  of  the  fraction  increases. 
Hence,  if  the  denominator  be  less  than  any  assignable  quantity, 
that  is  0,  the  value  of  the  fraction  will  be  greater  than  any  assign- 
able quantity,  that  is,  infinitely  great.  This  is  designated  by  the 
sign  go,  that  is,  c 

o=°°- 

TV.  When  c  is  o,  and  b  is  either  greater  or  less  than  a. 
If  we  put  a — b  equal  to  cZ,  then  35=-v 

We  are  now  to  determine  the  value  of  a  fraction  whose  numera- 
tor is  zero. 

Review. — 173.  When  is  the  value  of  x  positive?  When  negative? 
When  infinite?  Show  how  the  value  of  a  fraction  increases  as  its 
denominator  decreases.  Value  of  a  fraction  whose  denominator  is 
zero  ?     Of  £  when  c  is  0,  and  b  greater  or  less  than  a? 


148  RAY'S  ALGEBRA,  FIRST  BOOK. 

The  value  of  a  fraction  decreases  as  the  numerator  decreases. 
Hence,  if  the  numerator  be  less  than  any  assignable  quantity,  that 

is  0,  the  value  of  the  fraction  is  zero,  or  -^=0. 

V.  When  h=a,  and  c=0. 

In  this  case,  we  have  x= r— ?;• 

a — 6     0 

If  any  quantity  be  put  into  the  form  of  a  fraction,  and  both  terms 
be  divided  continually  by  the  same  quantity,  the  value  of  the  frac- 
tion will  remain  unchanged,  but  the  final  result  will  be  of  the 
form  g.  This  form  is  therefore  expressive  of  any  finite  value  what- 
ever.    Hence, 

We  say  that  &  is  the  symbol  of  indetermination ;  that  is,  the 
quantity  which  it  represents  has  no  particular  value. 

The  form  jl  sometimes  arises  from  a  particular  supposition,  when 
the    terms    of    a    fraction    contain    a    common    factor.      Thus,    if 

«2_&2                                                                                «2_ «2 
x= — ,   and  we  make  o=a,  it  reduces  to -=£;   but,  if 

we  cancel  a — 6,  and  then  make  6=a,  we  have  x=la.     Hence, 

Before  deciding  the  value  to  be  indeterminate,  we  must  see  that 
this  form  has  not  arisen  from  the  existence  of  a  factor  whose  value, 
by  a  particular  supposition,  is  zero. 

The  discussion  of  the  following  problem,  originally  pro- 
posed by  Clairaut,  will  serve  to  illustrate  further  the  pre- 
ceding principles. 


PROBLEM    OF    THE    COURIERS. 

Two  couriers  depart  at  the  same  time,  from  two  places, 
A  and  B,  distant  a  mi.  from  each  other ;  the  former  trav- 
els m  mi.  an  hour,  and  the  latter  n  mi.  j  where  will  they 
meet  ? 

There  are  two  cases  of  this  question. 

I.  When  the  couriers  travel  toward  each  other. 


Let  P  be  the  point  where  they  meet,      A 
and  «=AB,  the  distance  between  the 
two  places. 


TROBLEM  OF  THE  COURIERS.         149 

Let  x^AP,  the  distance  which  the  first  travels. 

Then,  a — a;— BP,  the  distance  which  the  second  travels. 

The  distance  each  travels,  divided  by  the  number  of  mi.  traveled 

in  1  hr.  will  give  the  number  of  hr.  he  was  traveling. 

x 
Therefore,  — =  the  number  of  hr.  the  first  travels. 
m 

And =  the  number  of  hr.  the  second  travels. 

n 
But  they  both  travel  the  same  number  of  hr.,  therefore, 

x     a — x 


m       n 
nx=am — mx 
am 


x- 


a — x- 


m-\-n 
an 


m-\-n 

1st.  Suppose  m=n-,  then,  £— -^—  =-»  and  a — X  =5-;  that  is,  if 
they  travel  at  the  same  rate,  each  travels  half  the  distance. 

2d.  Suppose  72=0;  then,  rc= —a;  that  is,  if  the  second  cou- 
rier remains  at  rest,  the  first  travels  the  whole  distance  from  A  to  B. 
These  results  correspond  to  the  circumstances  of  the  problem. 

II.  When  the  couriers  travel  in  the  same  direction. 


As  before,  let  P  be   the  point  of 


meeting,  each  traveling  in  that  direc-  B 

tion,  and  let  a=AB,  the  distance  between  the  places. 
a;=AP,  the  distance  the  first  travels. 
x — a=BP,  the  distance  the  second  travels. 
Then,  reasoning  as  in  the  first  case,  we  have 
x     x — a 
m~~    n 
nx=mx — am 

am 

x- 

x—a-. 

m — n 

1st.  If  we  suppose  m  greater  than  n,  the  value  of  X  will  be  pos- 
itive; that  is,  the  couriers  will  meet  on  the  right  of  B.  This  evi- 
dently corresponds  to  the  circumstances  of  the  problem. 


m—n 
an 


150  RAYS  ALGEBRA,  FIRST  BOOK. 

2d.  If  we  suppose  n  greater  than  m,  the  value  of  X,  and  also 
that  of  X — CL,  will  be  negative.  This  negative  value  of  x  shows  that 
the  point  of  meeting  is  to  the  left  of  A. 

Indeed,  when  on  is  less  than  ft,  it  is  evident  that  they  can  not 
meet,  since  the  forward  courier  is  traveling  faster  than  the  other. 
"We  may,  however,  suppose  that  they  had  met  previously. 

If  we  suppose  the  direction  in  which  the  couriers  travel  to  be 
changed;  that  is,  that  the  first  travels  from  A,  and  the  second 
from  B   toward  P/;   and    put  X=AY', 

a-\-x=BF/,  their  values  will  be  posi-    P/  [  ,  \  B 

five,  and  the  question  will  be  consist-  A- 

ent;  for  we  shall  then  have 

x      a-\-x 
rn       n 
am 


a+x= 


n — m 
an 


am       .  an 


3d.  If  we  suppose  m=n ;  then,  x=-q-,  and  x—a—-^-. 

These  values  of  X,  and  a — X,  being  equal  to  infinity,  Art.  173,  it 
follows  that  if  the  couriers  travel  at  the  same  rate,  the  one  can  never 
overtake  the  other.  This  is  sometimes  expressed  by  saying,  they  only 
meet  at  an  infinite  distance. 

4th.  If  we  suppose  a=0;  then,  x= ,  and  x — a= . 

Tit — IX  llh lh 

It  has  been  shown  already,  that  these  values  are  equal  to  0. 
Hence,  if  the  couriers  are  no  distance  apart,  they  will  have  to  travel 
no  (0)  distance  to  be  together. 

5th.  If  we  suppose  m=n,  and  a=0;  then,  z=g,  and  x—  a=g. 

It  has  already  been  shown  that  this  form  is  expressive  of  any 
finite  value  whatever.  Hence,  if  the  couriers  are  no  distance  apart, 
and  travel  at  the  same  rate,  they  will  be  always  together. 

Review. — 173.  What  is  the  value  of  x  when  b=a  and  c=0?  Of  a 
fraction  whose  terms  are  both  zero?  How  does  this  form  sometimes 
arise? 

173.  Discuss  the  problem  of  the  "Couriers,"  and  show,  that  in 
every  hypothesis  the  solution  corresponds  to  the  circumstances  of 
the  problem. 


IMPOSSIBLE  PROBLEMS.  151 


dill 
Lastly,  if  we  suppose  n=0;  then,  X= — =a;    that  is,  the  first 

courier  travels  from  A  to  B,  overtaking  the  second  at  B. 

7fl  2&7YI 

If  we  suppose  W=-c-;   then,  x=— — =2a,  and  the  first  travels 
twice  the  distance  from  A  to  B,  before  overtaking  the  second. 


CASES  OF  INDETERMINATION  IN  SIMPLE  EQUATIONS,  AND 
IMPOSSIBLE  PROBLEMS. 

174.  An  Independent  Equation  is  one  in  which  the 
relation  of  quantities  which  it  contains,  can  not  be  obtained 
directly  from  others  with  which  it  is  compared.  Thus,  the 
equations,  se-|-2y=ll 

2x+by=2Q 

are  independent  of  each  other,  since  the  one  can  not  be 
obtained  from  the  other  in  a  direct  manner.  But  the 
equations,  x-\-2y=ll 

2x-\-4:7/=22 

are  not  independent  of  each  other,  the  second  being  derived 
directly  from  the  first  by  multiplying  both  sides  by  2. 

175.  An  Indeterminate  Equation  is  one  that  can  be 
verified  by  different  values  of  the  same  unknown  quantity. 

Thus,    in    the    equation    x — 2/=5,    by    transposing   y,   we    have 

If  we  make  2/=l,  X=6.  If  we  now  make  £/=2,  SC=7,  and  so  on. 
from  which  it  is  evident  that  an  unlimited  number  of  values  may  be 
given  to  x  and  y.  that  will  verify  the  equation. 

If  we  have  two  equations  containing  three  unknown 
quantities,  we  may  eliminate  one  of  them ;  this  will  leave 
a  single  equation  containing  two  unknown  quantities,  which, 
as  in  the  preceding  example,  will  be  indeterminate.  Thus, 
if  we  have 

x+3y-\-z=10 ;  and 

X-\-2y — z=  6;  if  we  eliminate  x1  we  have 
y-\-2z—  4;  from  which  y=£— 2z. 


152  RAY'S  ALGEBRA,  FIRST  BOOK. 

Putting  z=l,    y=2,  and  x=10—3y—z=3. 
Putting  2=1  £,  2/=l,  and  x=5%]  and  so  on. 

Other  examples  might  be  given,  but  these  are  sufficient 
to  show,  that 

When  the  number  of  unknown  quantities  exceeds  the  num- 
ber of  independent  equations,  the  problem  is  indeterminate. 

A  question  is  sometimes  indeterminate  that  involves  only 
one  unknown  quantity ;  the  equation  deduced  from  the 
conditions,  being  of  that  class  denominated  identical ;  as 
the  following : 

"What  number  is  that,  of  which  the  f,  diminished  by 
the  §,  is  equal  to  the  j$  increased  by  the  -fat 

Let  X=  the  number. 

&x      2x      x        x 
Then,  T__  — +  _. 

Clearing  of  fractions,  45a; — 4tix=%x-{- 2x;  or,  5x=5x,  which  will 
be  verified  by  any  value  of  x  whatever. 

176.  The  reverse  of  the  preceding  case  requires  to  be 
considered ;  that  is,  when  the  number  of  equations  is 
greater  than  the  number  of  unknown  quantities.  Thus, 
we  may  have 

x+  y=10    (1.) 

x-  y=  4    (2.) 

2z— 3^=  5     (3.) 

Each  of  these  equations  being  independent  of  the  other 
two,  one  of  them  is  unnecessary,  since  the  values  of  x  and 
y,  which  are  7  and  3,  may  be  determined  from  any  two 
of  them. 

When  a  problem  contains  more  conditions  than  are  neces- 
sary for  determining  the  values  of  the  unknown  quantities, 
those  that  are  unnecessary  are  termed  redundant  conditions. 

The  number  of  equations  may  exceed  the  number  of  un- 
known quantities,  so  that  the  values  of  the  unknown  quail- 


IMPOSSIBLE  PROBLEMS.  153 

tities  shall  be  incompatible  with  each  other.     Thus,  if  we 
have 

*+  y=  9   (i.) 

x+<ly=A%     (2.) 
2x+3y=21     (3.) 

The  values  of  x  and  y,  found  from  equations  (1)  and  (2),  are 
x=5,  y=4:;  from  equations  (1)  and  (3),  are  #=6,  y=3\  and  from 
equations  (2)  and  (3),  are  x=3,  y=5.  From  this  it  is  manifest, 
that  only  two  of  these  equations  can  be  true  at  the  same  time. 

A  question  that  contains  only  one  unknown  quantity  is 
sometimes  impossible ;  as  the  following : 

What  number  is  that,  of  which  the  i  and  |  diminished 
by  4,  is  equal  to  the  |  increased  by  8  ? 

Let  X==  the  number;  then,  ^+0 — ^=z~aJr^- 

Clearing  of  fractions,  3#+2z— 24=5rc-f  48. 

By  subtracting  equals  from  each  side,  0=72,  which  shows  that 
the  question  is  absurd. 

ITT.  Take  the  equation  ox — cx=b — d,  in  which  a  rep- 
resents the  sum  of  the  positive,  and  — c  the  sum  of  the 
negative  coefficients  of  x ;  h  the  sum  of  the  positive,  and 
— d  the  sum  of  the  negative  known  quantities. 

This  will  evidently  express  a  simple  equation  involving  one  un- 
known quantity,  in  its  most  general  form. 
This  gives      (a— c)ar— &— d. 

Let  a—c=m,  and  b—d—n. 

71 
We  then  have  mx=n,  or  x=z— • 

m 

Now,  since  n  divided  by  m  can  give  but  one  quotient,  we  infer 
that  an  equation  of  the  first  degree  has  but  one  root;  that  is,  in  a  simple 
equation  involving  but  one  unknown  quantity,  there  is  but  one  value 
that  will  verify  the  equation. 

Review. — 174.  When  is  an  equation  termed  independent?  Ex- 
ample. 175.  When  said  to  be  indeterminate'?  Example.  176.  What 
are  redundant  conditions? 


154  RAY'S  ALGEBRA,  FIRST  BOOK. 


VI.    OF    POWERS,    ROOTS,    AND 
RADICALS. 

INVOLUTION,    OR    FORMATION    OF    POWERS. 

ITS.  The  Power  of  a  quantity  is  the  product  arising 
from  multiplying  the  quantity  by  itself  a  certain  number 
of  times. 

The  Boot  of  the  power  is  the  quantity  to  be  multiplied. 

Thus,  a2  is  called  the  second  power  of  a,  because  a  is 
taken  twice  as  a  factor  ;  and  a  is  called  the  second  root 
of  a\ 

So,  also,  of  is  the  third  power  of  a,  because  aX«X«=«s, 
a  being  taken  three  times  as  a  factor ;  and  a  is  the  third 
root  of  a3. 

The  second  power  is  generally  called  the  square,  and  the 
second  root,  the  square  root.  In  like  manner,  the  third 
power  is  called  the  cube,  and  the  third  root,  the  cube  root. 

The  Exponent  is  the  figure  indicating  the  power  to  which 
the  quantity  is  to  be  raised.  It  is  written  on  the  right,  and 
a  little  above  the  quantity.     See  Arts.  33  and  35. 


CASE    I. 

TO    RAISE    A    MONOMIAL    TO    ANY    GIVEN    POWER. 

179. — 1.  Let  it  be  required  to  raise  2ab2  to  the  third 
power. 

According  to  the  definition,  the  third  power  of  2ab2  will 
be  the  product  arising  from  taking  it  three  times  as  a  factor. 

Thus,     (2a&2)3=2a62x2a62x2a&2:=2x2x2aaa&2&2&2 

=23xa1+1+1X62+2+2=23X«lx3X62x3=8«366- 


FORMATION  OF  POWERS.  155 

The  coefficient  of  the  power  is  found  by  raising  the 
coefficient,  2,  of  the  root,  to  the  given  power ;  and  the  ex- 
ponent of  each  letter,  by  multiplying  the  exponent  of  the 
letter  in  the  root  by  3,  the  index  of  the  required  power. 

ISO.  "With  regard  to  the  signs  of  the  different  powers, 
there  are  two  cases. 

First,  when  the  root  is  positive;  and  second,  when  neg- 
ative. 

1st.  When  the  root  is  positive. 

Since  the  product  of  any  number  of  positive  factors  is 
always  positive,  it  is  evident  that  if  the  root  is  positive, 
all  the  powers  will  be  positive. 

Thus,  -f  «x+«=+«2 

-f  aX+«X+a=+a3;  and  so  on. 

2d.  When  the  root  is  negative. 

Let  us  examine  the  different  powers  of  a  negative  quan- 
tity, as  — a. 

— a=  first  power,  negative. 
— aX — a=-|-a2=  second  power,  positive. 
— aX — aX — a— — ^3=  third  power,  negative. 
— «X — aX — aX~ <3=-f  a4=  fourth  power,  positive. 

-aX — &X— aX— aX— a=— «5=  fifth  power,  negative. 

From  this  we  see  that  the  even  powers  of  a  negative 
quantity  are  positive,  and  the  odd  powers  negative.     Hence, 


TO    RAISE    A    MONOMIAL    TO    ANY    GIVEN    POWER, 

Rule. — 1.  Raise  the  numeral  coefficient  to  the  required 
power,  and  multiply  the  exponent  of  each  letter  by  the  ex- 
ponent of  the  power. 

2.  If  the  monomial  is  positive,  all  the  powers  will  be  posi- 
tive; if  negative,  the  even  powers  will  be  positive,  and  the 
odd  powers  negative. 


156 


RAY'S  ALGEBRA,  FIRST  BOOK. 


1. 

Find  the  square  of  3ax2y3. 

.     .    Ans.  9aVj/6. 

2. 

Square  of  bb2c3.    .     . 

.     .     Ans.  2bbic«. 

3. 

Cube  of  2x2xf 

.     .  Ans.  Sx6y9. 

4. 

Square  of  — ah2c.  .     .     . 

.     .     .Ans.  aW. 

5. 

Cube  of  — abc2.      .     .     . 

.      .  Ans.  — a3b3c6. 

6. 

Fourth  power  of  Sab3c2. 

.      Ans.  81a4612c8. 

7. 

Fourth  power  of  — Sab3c2. 

Ans.  Sla*b12t*. 

8. 

Fifth  power  of  ab3cd2.    .     . 

Ans.  a*b*<*dF. 

9. 

Fifth  power  of  — ab3cd2.     . 

Ans.  —  a5Z>15c5d10. 

10. 

Seventh  power  of  — m2n3. 

.Ans.  — mun11. 

11. 

Eighth  power  of  — mu2. 

.     .  Ans.  m8>ils. 

12. 

Cube  of  — Sxy2. 

Ans.  — 2>lx3if. 
Ans.  625a8#12. 

13. 

Fourth  power  of  ba2x3.  . 

14. 

Fourth  power  of  7«2#3.  . 

Ans.  2401aV2. 

15. 

Fifth  power  of  — 2>d2xy2z3. . 

A.ns.  —  24:3alQxy°z15. 

CASE    II. 


TO    RAISE    A    POLYNOMIAL    TO    ANY    POWER, 

181.  Rule. — Find  the  product  of  the  quantity,  taken  as 
a  factor  as  many  times  as  there  are  units  in  the  exponent  of 
the  power. 

1.  Find  the  square  of  ax-\-cy. 

Ans.  (a.x-\-cy) ^ax-\-cy^-=a2x2-\~2acxy-\-c2y'2. 


2.  Square  of  1 — x.    . 

3.  Square  of  x-\-l.    . 

4.  Square  of  2x2— 3/. 

5.  Cube  of  a-f- x.  . 

6.  Cube  of  x — y.  . 

7.  Cube  of  2x— 1.     . 


.  Ans.  1 — 2x-\-xi. 
....  Ans.  a2+2.T+l. 
.  Ans.  4x4—  I2x2y2+§y\ 
.  Ans.  a3-\-Sa2x-]-Sax2-\-x3. 
.  Ans.  x3 — 2>x2y-\-Zxy2 — y3. 
.   Ans.  Sx3— 12x2+6x— 1. 


Review. — 177.  Show  that  in  an  equation  of  the  first  degree,  the 
unknown  quantity  can  have  but  one  value.  178.  What  does  the 
term  power  denote?  The  term  root?  What  is  the  second  power 
of  a?  Why?  The  third  power  of  a?  Why?  What  is  the  second 
power  generally  called?  The  second  root?  What  is  the  exponent? 
Where  should  it  be  written? 


FORMATION  OF  POWERS. 


157 


8.  Find  the  square  of  a — b-\-c — d. 

Ans.  a2— 2ab  +  b2+2ac— 2ad+c2— 2bc-\-2bd—  2cd-{- d\ 

9.  Find  the  cube  of  2x2— 3x-f  1. 

Ans.  8x6— 36a5+66x4— 63*3+33a;2— 9*-fl. 


CASE     III. 
TO    RAISE    A    FRACTION    TO    ANY    POWER, 

X82«  Rule. — Raise  both  numerator  and  denominator  to 
the  required  power. 

a2+2ab-{-b2 


1.  Find  the  square  of 

2.  Find  the  square  of 


3.  Find  the  cube  of 


a+b 


c — d' 

2x 


3y      *     * 


xhj 


4.  Find  the  square  of  — 

5.  Find  the  square  of 


2tf_ 

w 

x—2 

z-f-3' 


6.  Find  the  cube  of  2a(£  /\ 

6yz2 

Ans. 


Ans. 


4x2 

w 

,3  ,.3 


.     .     Ans. 


Ans. 


a,c 


.     .     .     Ans. 
.  Ans 


4s4 

V 

a2— 4x-f  4 


a2-f6a;-f-9* 


ta3  (xz — ^x2y-\-  3xy2 — ?/) 


BINOMIAL     THEOREM. 

1S3.  The  Binomial  Theorem,  discovered  by  Sir  Isaac 
Newton,  explains  the  method  of  involving  a  binomial  quan- 
tity without  the  tedious  process  of  multiplication. 

To  illustrate  it,  we  shall  first,  by  means  of  multiplica- 
tion, find  the  different  powers  of  a  binomial. 

Review. — 179.  In  raising  2ab2  to  the  third  power,  how  is  the 
coefficient  of  the  power  found?  How  the  exponent  of  each  letter? 
180  When  the  root  is  positive,  what  is  the  sign  of  the  different 
j-jwers?     When  it  is  negative? 

180.  Rule  for  raising  a  monomial  to  any  given  power.  181.  A 
polynomial.  182.  A  fraction.  183.  What  does  the  Binomial  Theo- 
rem explain? 


158  RAYS  ALGEBRA,  FIRST  BOOK. 

1.  Let  us  first  raise  a-\-b  to  the  fifth  power. 
a  +  6 

a  -f  6 

a2-\-  a  b 
+  a  64-     62 

a24-2a  64-     b2= second  power  of  a+b,  or  (a-f-5)2. 

a  +  b 
a3+2a26+    a  62 

a26-f-  2a  62+     63 
a34-3a26+  3a  62+     63=    .    .     third  power  of  a+b,  or  (a+6)3. 
a  +6 
a4+3a364-  3a262+    a  63 

-f  a36+  3a2624-  3a  63+6* 

a4+4a36+  6a262+  4a  63-j-64= (a+b)*. 

a  -\-b 

a5+4a46+  6a3624-  4a2634-  ab* 

-f  a*b-\-  4a3624-  6a263-f4a64-f65 
a'4-5a)64-10a362+^10a263-f5a64-|-65= ,    .    .     .    .         .  (a-j-6)5. 

2.  Let  us  next  raise  a — b  to  the  fifth  power. 
a  —  b 

a  —  b 
a2—  a  b 

—  a  6+    62 

a2— 2a  64-    62=r (a— 6)2. 

a  —  6  N 

a3— 2a2&4-    a62 

_  a26-|-  2a  b2—    63 

a3_3a264-  3a  b2—    63_ (a— 6)3. 

a  —  6 

a4_3a36.|_  3a262—    a  63 

_  a364-  3a262_  3a  634_  &* 

a4_4a364-  6a262_  4a  634-  6*= (a— 6)*. 

a  —  6 

a^—ia^b-\-  6a362—  4a263~4-a64 

_  a46-f-  4a362—  6a263_|_4a64— 6* 
a*— 5a4  64-10a362_ 1 0a2634_5a64 — 6^= {a—bf. 

The  first  letter,  a,  is  called  the  leading  quantity;  and  the  second 
letter,  6,  the  following  quantity. 


FORMATION  OF  POWERS.  159 

184.  In  examining  these  examples,  we  discover  four 
laws,  as  follows : 

1st.  The  number  of  terms  of  the  power. 
2d.    The  signs  of  the  terms. 
3d.    The  exponents  of  the  letters. 
4th.  The  coefficients  of  the  terms. 

Let  us  examine  these  four  laws  separately. 

1st.  Of  the  Number  of  Terms.  As  the  second  power 
has  three  terms  ;  the  third  power,  four  terms ;  the  fourth, 
Jive  terms  ;  the  fifth,  six  terms ;  we  infer,  that 

The  number  of  terms  in  any  power  of  a  binomial  is  one 
greater  than  the  exponent  of  the  power. 

2d.  Of  the  Signs  of  the  Terms.  When  both  terms  are 
positive,  all  the  terms  will  be  positive. 

When  the  first  term  is  positive,  and  the  second  negative, 
'  the  od 

NEGATIVE. 


all  the  ODD   terms  will  be   POSITIVE,  and  the  EVEN  terms 


3d.  Of  the  Exponents  of  the  Letters.  If  we  omit  the 
coefficients,  the  remaining  parts  of  the  fifth  powers  of 
a-\-b  and  a — b,  are 

(a-\-bf a»-f-a*&-l-a3&2^a2&3_i_a&4_j_&5> 

(CL—bf a5_a4ft_|_a3&2_a2&3_|_a&4_&5> 

An  examination  of  these  and  the  other  powers  of  a-\-b 
and  a — b,  shows  that 

1 .  The  exponent  of  the  leading  letter  in  the  first  term  is 
the  same  as  that  of  the  power  of  the  binomial;  in  the  other 
terms,  it  decreases  by  unity  from  left  to  right,  and  disappears 
in  the  last  term. 

2.  Tlie  following  letter  begins  with  an  exponent  of  one,  in 
the  second  term;  increases  by  unity  from  left  to  right;  and, 
in  the  last  term,  is  the  same  as  the  power  of  the  binomial. 


1G0 


RAY'S  ALGEBRA,  FIRST  BOOK. 


3.  The  sum  of  the  exponents  of  the  two  letters  in  any  term 
is  always  the  same,  and  is  equal  to  the  power  of  the  bi- 
nomial. 


Write  the  different  powers   of  the  following 
without  the  coefficients  : 


binomials 


(*-H/)3 

{x—yy 
{x+yf 
{x—yf 

(x-y)7 
(x+yf 


x^x2y-\-xy2-\-yz 

x^—xhj-^x2y2—xy^yK 

x^xiy-^x^yZ-^xZyZJf-xy^y5. 

x5 — xr}y-\-x*y2 — x3y3-\-x2y4 — xy^>-\-y^. 

x7 — x(]y-{-xr,y2 — x^-\-x^yA — x2y:"^\-xij' — y7. 

x*-\-x7y^xCly2-\-xr>y5-\-x4yiJrx2iyrj-\-x2y(!>-\-xy7-\-y8. 


4th.  Of  the  Coefficients  of  the  Terms.  The  coefficient 
of  the  first  and  last  terms  is  always  1;  the  coefficient  of 
the  second  term  is  the  same  as  that  of  the  power  of  the 
binomial. 

The  law  of  the  other  coefficients  is  as  follows: 

If  the  coefficient  of  any  term  be  multiplied  by  the  exponent 
of  the  leading  letter,  and  the  product  be  divided  by  the  num- 
ber of  that  term  from  the  left,  the  quotient  will  be  the  coeffi- 
cient of  the  next  term. 

Omitting  the  coefficients,  the  terms  of  a-\-b  raised  to  the 
sixth  power,  are 

a6_|_a56-j-a462_j_a363_j_a26i_i_a65_|_56# 

The  coefficients,  according  to  the  above  principles,  are 
15x4     20x3 


6, 


6X5 

'  2   ' 
15, 


•6     •         4 
Or,   1,     6,      15,  20,  15, 

Hence,  {a+bf^at+Wb+lZaW^aW+VoaW+Qafr-tb*. 


15X2 
5  » 
6, 


6X1 
6  * 
1. 


Review. — 184.  In  examining  the  different  powers  of  a  binomial, 
how  many  laws  are  discovered?  What  is  the  number  of  terms  in 
any  power  of  a  binomial?     Examples. 

184.  When  both  terms  of  a  binomial  are  positive,  what  is  the  law 
of  the  signs?  When  one  term  is  positive,  and  the  other  negative? 
Give  the  law  of  the  exponents  in  its  three  divisions. 


FORMATION  OF  POWERS.  1G1 

From  this,  we  see,  that  the  coefficients  of  the  following 
terms  are  equal:  the  first  and  the  last;  the  second  from 
the  first,  and  the  second  from  the  last ;  the  third  from  the 
first,  and  the  third  from  the  last ;  and  so  on. 

Hence,  it  is  only  necessary  to  find  the  coefficients  of  half 
the  terms,  when  their  number  is  even,  or  one  more  than 
half,  when  odd ;  the  others  being  equal  to  those  already 
found. 

1.  Haise  x-\-y  to  tne  third  power.  Ans.  x3-{-Sx2y-{-Sxy2-{-y3. 

2.  Raise  (x — y)  to  the  fourth  power. 

Ans.  x* — 4x3y-\-6x2y2 — 4xy3-\-y*. 

3.  Raise  m-\-n  to  the  fifth  power. 

Ans.  m5-f5m%-|-10m3ft2  ^-10m2n3-j-5m7i4-j-?i5. 

4.  What  is  the  sixth  power  of  x — z  ? 

Ans.  x6— 6x5z-f-l  5»V— 20x3z*-\-15x2zi—  6xz5-{- z\ 

5.  What  is  the  seventh  power  of  a-\-b? 

Ans.  aY-f-7a65  +  21a5^-f35a463-f35a364+21a265-j-7G6H^. 

6.  What  is  the  eighth  power  of  m — ?i?  Ans.  m8 — 8m7n 
-|-28mV— 56?n5>i3-f  70mV—  56mV+28m2w6—  8m/i7-fw8. 

Y.  Find  the  ninth  power  of  x — y.     Ans.  x9 — Qxiy-\-2>Qxly'2 
— 84x6y-f-126^y— 126xy-f84x3/—  S6x2f-{-9xy&— y\ 

8.  Find  the  tenth  power  of  a-\-b. 

Ans.  a10+10a9Z>+45a8&2+120a7Z>3+210a6Z>4+252a5&5 
-f210a^6+120a3^+45a2Z>8-hl0aZ>94-610. 

185,  The  Binomial  Theorem  may  be  used  to  find  the 
different  powers  of  a  binomial,  when  one  or  both  terms 
consist  of  two  or  more  factors. 

Review. — 184.  To  what  is  the  coefficient  of  the  first  and  last 
terms  equal  ? 

J«H    Of  the   second   term?     How  is   the  coefficient  of  any  other 
term  found  ?     What  terms  have  their  coefficients  equal? 
1st  Bk.  14 


102  RAY'S  ALGEBRA,  FIRST  BOOK. 

1.  Find  the  cube  of  2x — ac2. 

Let  2x=m,  and  ac2=n ;  then,  2x — ac2=m — n. 
{m— n)3=rm3-  3m27i-f3mn2— n3 
m  —2x  n  z=a  c2 

m2=4x2  n2=dW 

w3=^8a;3  w3=a3c6 

Substituting  these  values  of  the  different  powers  of  m  and  n  in 
the  equation  above,  and  we  have 

(2x— ac2)3=8x*— 3x4o;2X«c2+3x2a:X«2c4— «3c6 
=&c3-l  2ac2z2+ 6a2c%— a3c6. 

2.  Find  the  cube  of  2a— 3b. 

Ads.  8a3— 36a2o  +  54a&2— 27&3. 

3.  Find  the  fourth  power  of  m-j-2n. 

Ans.  m4+8m37i-f24m2n2-f32?rf-j-16?i4. 

4.  Find  the  third  power  of  4ax'2-\-Scy. 

Ans.  64a3a;64-144a2c^4<y+108ac2a;y+27cy. 

5.  Find  the  fourth  power  of  2x — bz. 

Ans.  16x*— 160xh+600x2z2— 1000^3+625^4. 

18G.  The  Binomial  Theorem  may  likewise  be  used  to 
raise  a  trinomial  or  quadrinomial  to  any  power,  thus : 

1.  Find  the  second  power  of  a-\-b-\-c. 

Let  b-\-c=x\  then,  a+&-f-c=a+a:. 

(a-\-x)2=a2+2ax+x2 

2ax=2a(b+c)=2ab-\-2ae 
x2=(b+c)2=b2+2bc-\-e2 
Then,  (a+b-\-c)2=a2-{-2ab+2ac-\-b2-)-2bc-{-c2. 

2.  Find  the  third  power  of  x-\-y-\-z. 

Ans.  a?-\-  Sx2y -+- Sx2z-\-  2>xy2-\- 6xyz-{-  3xz2-\-y*-\-  Sy2z 
+  Syz2+z*. 

3.  Find  the  second  power  of  a-\-b-\-c-\-d. 

Ans.  a?-\-2ab+b2+2ac+2bc-{-c2+2ad+2bd+2cd+d2. 


EXTRACTION  OF  THE  SQUARE  ROOT.       163 

EVOLUTION. 

EXTRACTION  OF  THE  SQUARE  ROOT  OF  NUMBERS. 

187.  Evolution  is  the  process  of  finding  the  root  of  a 
quantity. 

The  Second,  or  Square  Root  of  a  number,  is  that  num- 
ber which,  being  multiplied  by  itself,  will  produce  the 
given  number. 

Thus,  2  is  the  square  root  of  4,  because  2x2=4. 

The  Extraction  of  the  Square  Root  is  the  process  of 
finding  the  second  root  of  a  given  number. 

188.  The  first  ten  numbers  and  their  squares  are 
1,      2,      3,      4,       5,      6,      7,      8,      9,      10, 

1,      4,      9,     16,     25,    36,     49,     64,     81,     100. 

The  numbers  in  the  first  line  are  also  the  square  roots 
of  the  numbers  in  the  second. 

Observing  the  ten  numbers  written  above,  we  see  that 
when  the  number  of  places  of  figures  in  a  number  is  not  more 
than  TWO,  the  number  of  places  of  figures  in  the  square  root 
will  be  ONE. 

Again,  take  the  following  numbers  and  their  squares : 
10,      20,      30,        40,        50,        60,        70,        80,        90,      100, 
100,     400,     900,     1600,    2500,     3600,     4900,     6400,     8100,  10000. 

From  this  we  see,  that  when  the  number  of  places  of 
figures  is  more  than  TWO,  and  not  more  than  FOUR,  the  num- 
ber of  places  of  figures  in  the  square  root  will  be  TWO. 

In  the  same  manner,  it  may  be  shown,  that  when  the 
number  of  places  of  figures  is  more  than  four,  and  not  more 
th^ii  :ix,  the  number  of  places  in  the  square  root  will  be 
three,  and  so  on.     Or,  thus ; 


164  RAYS  ALGEBRA,  FIRST  BOOK. 

When  the  number  of  places  of  figures  in  the  number  is 
either  one  or  two,  there  will  be  one  figure  in  the  root ; 

When  the  number  of  places  is  either  three  or  four,  there 
will  be  two  figures  in  the  root ; 

When  the  number  of  places  is  either  Jive  or  six,  there 
will  be  three  figures  in  the  root ;  and  so  on. 

ISO.  Every  number  may  be  regarded  as  being  com- 
posed of  tens  and  units.  Thus,  23  consists  of  2  tens 
and  3  units  ;  256  consists  of  25  tens  and  6  units.  There- 
fore, if  we  represent  the  tens  by  t,  and  the  units  by  u,  any 
number  will  be  represented  by  t-\-u,  and  its  square,  by  the 
square  of  t-{-u,  or  (t-\-ii)2. 

(t+uy=?+2tu+2?=t2+(2t+u-)u. 

Hence,  the  square  of  any  number  is  composed  of  the  square 
&f  the  tens,  plus  a  quantity  consisting  of  twice  the  tens  plus 
the  units,  multiplied  by  the  units. 

Thus,  the  square  of  23,  which  is  equal  to  2  tens  and  3  units,  is 

2  tens  squared  =(20)2=400 
(Twice  2  tens  -f  3  units)  X3=(40-|-3)X3=129 

529 

1.  Let  it  be  required  to  extract  the  square  root  of  529. 

Since  the  number  consists  of  three  places  529,23 

of  figures,  its  root  will  consist  of  two  places,  400| 

according  to  the   principles  in  Art.  188;    we  20x2=40129 
therefore   separate   it  into  two  periods,  as  in  3 

the  margin.  43  129 

Review. — 187.  What  is  the  square  root  of  a  number  ?  Example. 
188.  When  a  number  consists  of  only  one  figure,  what  is  the  great- 
est number  of  figures  in  its  square?  Examples.  When  a  number 
consists  of  two  places  of  figures?     Examples. 

188.  What  relation  exists  between  the  number  of  places  of  figures 
in  any  number  and  the  number  of  places  in  its  square?  189.  Of 
what  may  every  number  be  regarded  as  being  composed?  Prove 
this,  and  then  illustrate  it. 


EXTRACTION  OF  THE  SQUARE  ROOT.      165 

Since  the  square  of  2  tens  is  400,  and  of  3  tens,  900,  it  is  evident 
that  the  greatest  square  contained  in  500,  is  the  square  of  2  tens 
(20);  the  square  of  2  tens  (20)  is  400;  subtracting  this  from  529,  the 
remainder  is  129. 

Now,  according  to  the  preceding  theorem,  this  number  129  con- 
sists of  twice  the  tens  plus  the  units,  multiplied  by  the  units;  that 
is,  by  the  formula,  it  is  (2t-\-u)u. 

The  product  of  the  tens  by  the  units  can  not  give  a  product  less 
than  tens;  therefore,  the  unit's  figure  (9)  forms  no  part  of  the 
double  product  of  the  tens  by  the  units.  Then,  if  we  divide  the 
remaining  figures  (12)  by  the  double  of  the  tens,  the  quotient  will 
be  (3)  the  unit's  figure,  or  a  figure  greater  than  it. 

We  then  double  the  tens,  and  add  to  it  the  unit  figure  (3),  mak- 
ing 40-f-3=43  (2£-f-w);  multiplying  this  by  3  (w),  the  product 
is  129,  which  is  the  double  of  the  tens  plus  the  units,  multiplied  by 
the  units.  As  there  is  nothing  left  after  subtracting  this  from  the 
first  remainder,  we  conclude  that  23  is  the  exact  square 
root  of  529.  529J23 

In  squaring  the  tens,  and  also  in  doubling  them,  it  is  4 

customary  to  omit  the  ciphers,  though  they  are  under-       43il29 
stood.     Also,  the  unit's  figure  is  added   to  the  double  |129 

of  the  tens,  by  merely  writing  it  in   the  unit's  place. 
The  actual  operation  is  usually  performed  as  in  the  margin. 

2.  Let  it  be  required  to  extract  the  square  root  of  55225. 

Since  this  number  consists  of  five  places  of  figures,  its  root  will 
consist  of  three  places,  according  to  the  principles  in 
Art.  188;  we  therefore  separate  it  into  three  periods.  55225|235 

In  performing  this  operation,  we  find  the  square  4 

root  of  the  number  552,  on  the  same  principle  as  in     43jl52 
the  preceding  example.     We  next  consider  the  23  as  129 

so  many  tens,  and  proceed  to  find  the  unit's  figure  465|2325 
(5)  in  the  same  manner  as  in  the  preceding  example.  2325 

Hence, 


TO   EXTRACT   THE  SQUARE    ROOT    OF   WHOLE    NUMBERS, 

Rule. — 1.   Separate  the  given  numbers  into  periods  of  two 
places  each,  beginning  at  the  unit's  place. 


166  RAY'S  ALGEBRA,  FIRST  BOOK. 

2.  Find  the  greatest  square  in  the  left  period,  and  place  its 
root  on  the  right,  after  the  manner  of  a  quotient  in  division. 
Subtract  the  square  of  the  root  from  the  left  period,  and  to 
the  remainder  bring  down  the  next  period  for  a  dividend. 

3.  Double  the  root  already  found,  and  place  it  on  the  left 
for  a  divisor.  Find  how  many  times  the  divisor  is  contained 
in  the  dividend,  exclusive  of  the  right  hand  figure,  and  place 
the  figure  in  the  root,  and  also  on  the  right  of  the  divisor. 

4.  Multiply  the  divisor,  thus  increased,  by  the  last  figure  of 
the  root;  subtract  the  product  from  the  dividend,  and  to  the 
remainder  bring  down  the  next  period  for  a  new  dividend. 

5.  Double  the  whole  root  already  found  for  a  new  divisor, 
and  continue  the  operation  as  before,  until  all  the  periods  are 
brought  down. 

Note. — If,  in  any  case,  the  dividend  will  not  contain  the  divisor, 
the  right  hand  figure  of  the  former  being  omitted,  place  a  cipher  in 
the  root,  also  at  the  right  of  the  divisor,  and  bring  down  the  next 
period. 

lOO.  In  division,  when  the  remainder  is  greater  than 
the  divisor,  the  last  quotient  figure  may  be  increased  by 
at  least  1  ;  but  in  extracting  the  square  root,  the  remain- 
der may  sometimes  be  greater  than  the  last  divisor,  while 
the  last  figure  of  the  root  can  not  be  increased. 

To  know  when  any  figure  may  be  increased,  we  must 
determine  the  relation  between  the  squares  of  two  consecu- 
tive numbers. 

Let  a  and  a-\-l  be  two  consecutive  numbers. 
Then,   (a-hl)2— a2-f-2a-f-l,  is  the  square  of  the  greater. 
(a)2=a2  is  the  square  of  the  less. 

Their  difference  is      2a-f-l.     Therefore, 

Review. — 189.  Extract  the  square  root  of  529,  and  show  the  rea- 
son for  each  step,  by  referring  to  the  formula. 


EXTRACTION  OF  THE  SQUARE  ROOT.      167 

Tlie  difference  of  the  squares  of  two  consecutive  numbers  is 
equal  to  twice  the  less  number  increased  by  unity.     Hence, 

When  the  remainder  is  less  than  twice  the  part  of  the 
root  already  found,  plus  unity,  the  last  figure  can  not  be 
increased. 

Extract  the  square  root  of  the  following  numbers : 

950625.  .  Ans.  975. 
1525225.  .  Ans.  1235. 
412252416.  A.  20304. 

OF    THE    SQUARE    ROOT    OF    FRACTIONS. 
191.  Since  |X|=|j  therefore,  the  square  root  of  j  is  §; 


I 

4225.  . 

.  .  Ans.  65. 

4 

2. 

289444. 

.  Ans.  538. 

5 

0. 

498436. 

Ans.  706. 

6 

is>  n/sHtK    Hence> 


that  ^ 

l/9      d 

The  square  root  of  a  fraction  is  found  by  extracting  the 

square  root  of  both  terms. 

Before  extracting  the  square  root  of  a  fraction,  it  should 
be  reduced  to  its  lowest  terms,  unless  both  numerator  and 
denominator  are  perfect  squares.  The  reason  for  this  will 
be  seen  by  the  following  example : 

Find  the  square  root  of  J|. 

12     4x3 
Here,  qr=q — o      Now,  neither  12  nor  27  is  a  perfect 

square;    but,  by  canceling  the  common  factor  3,  the  frac- 
tion becomes  |?  the  square  root  of  which  is  f . 

When  both  terms  are  perfect  squares,  and  contain  a  com- 
mon factor,  the  reduction  may  be  made  either  before  or 
after  the  square  root  is  extracted. 

Review. — 189.  What  is  the  rule  for  extracting  the  square  root  of 
numbers?  190.  What  is  the  diiference  between  the  squares  of  two 
consecutive  numbers?  When  may  any  figure  of  the  quotient  be  in- 
creased? 


168  RAYS  ALGEBRA,  FIRST  BOOK. 

Thn<5     ,/lf) 4 2.    nr      16 4      nnr|    _/4 2 

±nus,  |/3S — g — 3  ,  or,   gg — y,  ana  y  ■$ — 5. 
Find  the  square  root  of  the  following  fractions 


1.  #j.     .     .     .  Ans.  &. 

2.  AY     .     .      .  Ans.  ,V 

3.  }|U.  .      .     .     Ans.  f. 


4.  #&.     .  .     .       Ans.  §. 

ft      _136  9  An«5       3  7 

u-     I  0000*    '  *        "  ^-us-    Too* 

ft-'     18225  Ans      135 

u*    roooooo*  •      ^u&-  TOOO' 


1 02.  A  Perfect  Square  is  a  number  whose  square  root 
can  be  exactly  ascertained;  as,  4,  9,  16,  etc. 

An  Imperfect  Square  is  a  number  whose  square  root 
can  not  be  exactly  ascertained;  as,  2,  3,  5,  6,  etc. 

Since  the  difference  of  two  consecutive  square  num- 
bers a2  and  a2-f-2a-f-l,  is  2a+l  ;  therefore,  there  are  al- 
ways 2a  imperfect  squares  between  them.  Thus,  between 
the  square  of  4(16),  and  the  square  of  5(25),  there 
are  8(2a=2x4)  imperfect  squares. 

A  Surd  is  a  root  which  can  not  be  exactly  expressed. 
Thus,  -j/2  is  a  surd  ;  it  is  1.414-f. 

The  signs  +  and  —  are  sometimes  placed  after  an  ap- 
proximate root,  to  denote  that  it  is  less  or  greater  than  the 
true  root. 

It  might  be  supposed,  that  when  the  square  root  of  a 
whole  number  can  not  be  expressed  by  a  whole  number, 
it  exactly  equals  some  fraction.  We  will  therefore  show, 
that 

The  square  root  of  an  imperfect  square  can  not  be  a  fraction. 

Let  c  be  an  imperfect  square,  such  as  2,  and,  if  possible, 
let  its  square  root  be  equal  to  a  fraction,  ^,  which  is  sup- 
posed to  be  in  its  lowest  terms. 

a  a? 

Then,  v/c=T  ;  and  c=— ,  by  squaring  both  sides. 
b  b 

Now,  by  supposition,  a  and  b  have  no  common  factor; 


EXTRACTION  OF  THE  SQUARE  ROOT.      169 

therefore,  their  squares,  a2  and   b2,  can  have   no   common 
factor,  since  to  square  a  number,  we  merely  repeat  its  fac- 

a2 
tors.      Consequently,  —  must  be  in  its  lowest  terms,  and 

can  not  be  equal  to  a  whole  number.     Therefore,  the  equa- 
tions c=^-,  and  i/c=7  are  not  true.     Hence, 
b2  b 

The  square  root  of  an  imperfect  square  can  not  be  a  fraction. 


APPROXIMATE    SQUARE    ROOTS. 

103.  To  illustrate  the  method  of  finding  the  approxi- 
mate square  root  of  an  imperfect  square,  let  it  be  required 
to  find  the  square  root  of  2  to  within  j. 

Reducing  2  to  a  fraction  whose  denominator  is  9  (the  square 
of  3,  the  denominator  of  the  fraction  |),  we  have  2=*-£. 

The  square  root  of  18  is  greater  than  4,  and  less  than  5;  and  the 
square  root  of  ^  is  greater  than  |,  and  less  than  | ;  therefore,  |  is 
the  square  root  of  2  to  within  less  than  I.     Hence, 


TO    EXTRACT   THE    SQUARE    ROOT    OF  A  WHOLE    NUMBER  TO 
WITHIN    A    GIVEN    FRACTION, 

Rule. — 1.  Multiply  the  given  number  by  the  square  of  the 
denominator  of  the  fraction  which  determines  the  degree  of 
approximation, 

2.  Extract  the  square  root  of  this  product  to  the  nearest 
unit,  and  divide  the  result  by  the  denominator  of  the  fraction. 

Review. — 191.  How  is  the  square  root  of  a  fraction  found,  when 
both  terms  are  perfect  squares  ?  192.  When  is  a  number  a  perfect 
square?  Examples.  When  an  imperfect  square?  How  determine 
the  number  of  imperfect  squares  between  any  two  consecutive  per- 
fect squares? 

192.  What  is  a  root  called,  which  can  not  be  exactly  expressed? 
Provp  that  the  square  root  of  an  imperfect  square  can  not  be  a  frac- 
tion. 193.  How  find  the  approximate  square  root  of  an  imperfect 
square  to  within  any  given  fraction?  How,  when  the  fraction  is  a 
decimal? 

1st  Bk.  15* 


170  RAYS  ALGEBRA,  FIRST  BOOK. 

1.  Find  the  square  root  of  5  to  within  i.     .     Ans.  2l. 

2.  Of     7  to  within  ^ .  Ans.  2T%. 

3.  Of  27  to  within  ^ Ans.  5j. 

4.  Of  14  to  within  TL Ans.  3.7. 

5.  Of  15  to  within  ,£, Ans.  3.87. 


As  the  squares  of  10,  100,  etc.,  are  100,  10000,  etc., 
the  number  of  ciphers  in  the  square  of  the  denominator 
of  a  decimal  fraction  equals  twice  the  number  in  the  de- 
nominator itself.     Therefore, 

When  the  fraction  which  determines  the  degree  of  approxi- 
mation is  a  decimal,  add  two  ciphers  for  each  decimal  place 
required,  extract  the  root,  and  point  off  from  the  right,  one 
place  of  decimals  for  each  two  ciphers  added, 

6.  Find  the  square  root  of  2  to  six  places  of  decimals. 

Ans.  1.414213. 

7.  Find  the  square  root  of     10.  Ans.  3.162277+. 

8.  Find  the  square  root  of  101.        Ans.  10.049875-f. 


19  1.  To  find  the  approximate  square  root  of  a  fraction. 

within  7j, 


1.  Let  it  be  required   to   find  the  square  root  of  | 
dthin  i. 

3 3Yf 2  1 

7 7*7 4<J- 


Now,  since  the  square  root  of  21  is  greater  than  4,  and 
less  than  5,  the  square  root  of  J  J  is  greater  than  ^,  and 
less  than  | ;  therefore,  4  is  the  square  root  of  ^  to  within 
less  than  ^.     Hence, 

If  we  multiply  the  numerator  of  a  fraction  by  its  denomi- 
nator, then  extract  the  square  root  of  the  product  to  the  near- 
est unit,  and  divide  the  result  by  the  denominator,  the  quotient 
will  be  the  square  root  of  the  fraction  to  within  one  of  its 
equal  parts. 


EXTRACTION  OF  THE  SQUARE  ROOT.       171 

2.  Find  the  square  root  of  T4T  to  within  -fa.         Ans.  -^. 

3.  Find  the  square  root  of  -fa  to  within  y1-.  Ans.  |. 

Since  any  decimal  may  be  written  in  the  form  of  a  frac- 
tion having  a  denominator  a  perfect  square,  by  adding 
ciphers  to  both  terms  (thus,  .4=-^=-^$$,  etc.),  there- 
fore, as  shown  in  Art.  193,  its  square  root  may  be  found 
by  the  following 

Rule. — Annex  ciphers,  until  the  number  of  decimal  places 
shall  be  double  the  number  required  in  the  root,  extract  the 
root,  and  point  off  from  the  right  the  required  number  of 
decimal  places. 

Find  the  square  root 

4.  Of     .6  to  six  places  of  decimals.         Ans.  .,7f74596. 

5.  Of  .29  to  six  places  of  decimals.         Ans.   .538516. 

The  square  root  of  a  whole  number  and  a  decimal  may 
be  fcund  in  the  same  manner.     Thus, 

The  square  root  of  2.5  is  the  same  as  the  square  root 
of  £go,  which,  carried  out  to  6  places  of  decimals,  is 
1.581138+. 

6.  Find  the  square  root  of  10.76  to  six  places  of  deci- 
mals. Ans.  3.280243. 

7.  Find  the  square  root  of  1.1025.     .     .     .    Ans.  1.05. 

"When  the  denominator  of  a  fraction  is  a  perfect  square, 
extract  the  square  root  of  the  numerator,  and  divide  the 
result  by  the  square  root  of  the  denominator ;  or,  reduce 
the  fraction  to  a  decimal,  and  then  extract  its  square  root. 

R/'-tew. — 194.  How  find  the  approximate  square  root  of  a  frac- 
tion to  within  one  of  its  equal  parts?  How  extract  the  square  root 
of  a  decimal?  Of  a  fraction,  when  both  terms  are  not  perfect 
squares? 


172  RAYS  ALGEBRA,  FIRST  BOOK. 

When  the  denominator  of  the  fraction  is  not  a  perfect 
square,  the  latter  method  should  be  used. 

8.  Find  the  square  root  of  |  to  five  places  of  decimals. 
j/3=1.73205+,  ^4=2,  .•.1/f^=iJ^aw+=.86602-f . 

Or,  |=.T5.  and  1/775~=:.86602+. 

9.  Find  the  square  root  of  8$.  Ans.  1.795054+. 

10.  Find  the  square  root  of  ^.  Ans.  .661437  +  . 

11.  Find  the  square  root  of  5|.  Ans.  2.426703-}-. 

12.  Find  the  square  root  of  |.  Ans.  .377964+. 

SQUARE    ROOT    OF    MONOMIALS. 

10*5.  To  square  a  monomial,  Art.  179,  we  square  its 
coefficient,  and  multiply  the  exponent  of  each  letter  by  2. 
Thus, 


(Sab2y=9a2V.     Therefore,  l/9a2bi=Sab\     Hence, 

TO  EXTRACT  THE  SQUARE  ROOT  OF  A  MONOMIAL, 

Rule. — Extract  the  square  root  of  the  coefficient,  and  divide 
the  exponent  of  each  letter  by  2. 

Since  +aX+^=+a25  and  — aX — a=-{-a2, 
Therefore,  j/a2=+a,  or  — a. 

Hence,  the  square  root  of  any  positive  quantity  is  either 
phis  or  minus.  This  is  generally  expressed  by  using  the 
double  sign.  Thus,  -j/4a2 — J_2a,  which  is  read,  plus  or 
minus  2a. 

If  a  monomial  is  negative,  the  extraction  of  the  square 
root  is  impossible,  since  the  square  of  any  quantity,  either 
positive  or  negative,  is  necessarily  positive.  Thus,  j/ — 9, 
■j/ — 4a2,  j/ — b,  are  algebraic  symbols,  which  indicate  im- 
possible operations. 


EXTRACTION  OF  THE  SQUARE  ROOT.       173 

Such,  expressions  are  termed  imaginary  quantities.  When 
they  result  from  an  equation  or  a  problem,  they  indicate 
some  absurdity  or  impossibility.     See  Art.  218. 

Find  the  square  root  of  the  following  monomials: 


1.  4a2x\.    .      .    Ans.  ±2ax. 

2.  9xy.    .      .    Ans.  rb3xy2. 

3.  36a*t8a;2.    Ans.  ±6a2l3x. 


4.  49a2Z>V.     Ans.  dtzIabW. 

5.  625zV.      Ans.  ±25a*2. 

6.  1156a2xV\  A.  ±34okbV. 


Since  (  -  )  =  ?-,,  therefore,  yj— ==±=v  I  hence,  to  find  the 
square  root  of  a  fraction,  extract  the  square  root  of  both 
terms. 

1 6x2i/*  4xy2 

7.  Find  the  square  root  of       *  '  _.      .     .     .    Ans.±-^ — - 


SQUARE    ROOT    OF    POLYNOMIALS. 

196.  In  order  to  deduce  a  rule  for  extracting  the 
square  root  of  a  polynomial,  let  us  first  examine  the  rela- 
tion that  exists  between  the  several  terms  of  any  quantity 
and  its  square. 

(a+b)2=a2+2ab+b2=a2+(2a+b)b. 

(a+b+c)2  =  a2+2a&+&2-f-2ac+26c+c2  =  a2-f  (2a+6)6+(2a 
_}_26+c)c. 

(a-f  6+e-j-d)2  =  a2+2a6+62+2ac+26c+c2+2ac?+26d+2crf 
-f-d2=a2+(2a-f6)6+(2a+26+c)c+(2a+26+2c+<i)c?. 

Hence,  the  square  of  any  polynomial  is  formed  accord- 
ing to  the  following  law : 

The  square  of  any  polynomial  is  equal  to  the  square  of  the 
first  term — plus  twice  the  first  term,  plus  the  second,  multi- 

Review.— 195.  How  find  the  square  of  a  monomial?  How  find  its 
square  root?  What  is  the  sign  of  the  square  root  of  any  positive 
quantity? 

195.  Why  is  the  extraction  of  the  square  root  of  a  negative  mono- 
mial impossible?  Give  examples  of  symbols  that  indicate  impos- 
sible operations.     What  are  they  termed?     What  do  they  indicate? 


174  KAYS  ALGEBRA,  FIRST  BOOK. 

plied  by  the  second — plus  twice  the  first  and  second  terms, 
plus  the  third,  multiplied  by  the  third — plus  twice  the  first, 
second,  and  third  terms,  plus  the  fourth,  midtiplied  by  the 
fourth;  and  so  on. 

Hence,  by  reversing  the  operation,  we  have  the  following 


RULE, 
FOR    EXTRACTING    THE    SQUARE    ROOT    OF  A    POLYNOMIAL. 

1.  Arrange  the  polynomial  with  reference  to  a  certain  letter. 

2.  Extract  the  square  root  of  the  first  term,  place  the  result 
on  the  right,  and  subtract  its  square  from  the  given  quantity. 

3.  Divide  the  first  term  of  the  remainder  by  double  the 
root  already  found,  and  annex  the  result  both  to  the  root  and 
the  divisor.  Multiply  the  divisor  thus  increased,  by  the  second 
term  of  the  root,  and  subtract  the  product  from  the  remainder. 

4.  Double  the  terms  of  the  root  already  found,  for  a,  par- 
tial divisor,  and  divide  the  first  term  of  the  remainder  by  the 
first  term  of  the  divisor,  and  annex  the  result  both  to  the  root 
and  the  partial  divisor.  Multiply  the  divisor  thus  increased, 
by  the  third  term  of  the  root,  and  subtract  the  product  from 
the  last  remainder. 

5.  Proceed  in  a  similar  manner  until  the  work  is  finished. 

Remark. — If  the  first  term  of  any  remainder  is  not  exactly 
divisible  by  double  the  first  term  of  the  root,  the  polynomial  is  not  a 
perfect  square. 

1.  Find  the  square  root  of  r2-f-2rr/+r'2+2rr"-^2r//,-{-/,2. 
ri+2rr/-\-r/2Jr2rr"-ir2r/r"-\-r"2\r+r'-irr",  root. 


2r-\-r' 


2rr'4-r/2 
2rr/4-r/2 


2r+2r/-\-r// 


2rr//-\-2r/r//-\-r"2 
2rr//4-2r/ry/4-r//2 


EXTRACTION  OF  THE  SQUARE  ROOT.      175 

The  square  root  of  the  first  term  is  r,  which  write  as  the  first  term 
of  the  root.  Subtract  the  square  of  r  from  the  given  polynomial, 
and  dividing  the  first  term  of  the  remainder  2rr/,  by  2r,  the  double 
of  the  first  term  of  the  root,  the  quotient  is  r/,  the  second  term  of 
the  root. 

Next,  place  r*  in  the  root,  and  also  in  the  divisor,  and  multiply 
the  divisor, thus  increased, by  r\  and  subtract  the  product  from  the 
first  remainder. 

Double  the  terms  r-\-rf,  of  the  root  already  found,  and  proceed  as 
before,  until  the  work  is  finished. 

2.  Find  the  square  root  of  2$>xhf—24xif— \2xhj+\x<> 
+  16/. 

Arranging  the  polynomial  with  reference  to  x,  we  have 


4x*— 12x*y-^25x2y2— 24^+16^ [2a;2— 3xy+4y2,  root, 
4^4 


4a;2 — 3xy 


—\2xHj\2hx2y2 
— 12a%+  9x2y2 


4x2 — 6xy+4y2 


I6x2y2— 24a#3+16&/* 
I6x2y2— 24xy*+l  6?/4 


Find  the  square  root  of  the  following  polynomials : 

3.  x2+4x-\-± Ans.  x-\-2. 

4.  x2y2—8xy-\-16 Ans.  xy— 4. 

5.  4:a2x?-\-2by2z2 — 20axyz Ans.  2ax — hyz. 

6.  cc4+4x3+6x2+4a;+l Ans.  x2-\-2x+l. 

7.  9/— 12f+S4y2— 20y+25.      .      Ans.  3^—2^+5. 

8.  1— 4x-f-10a;2— 20x3+25x4— 24z5+16a6. 

Ans.  1— 2x-\-3x2—  4x*. 

9.  a6— Qa'x+lSa'x2— 20a3x3+15aV—  6ax5+x\ 

Ans.  a3 — 3a2x-\-3ax2 — x3. 

10.  x2-\-ax-\-\a2 Ans.  x-\-la. 

11.  x2— 2x+l  +  2xy— 2y+y2.     .     .      .      Ans.  x+y—l. 

12.  <r/r+l)(a.+  2Xaj+3)  +  l.     .     .     .  Ans.  x2+3x+l. 

Review.— 196.  What  is  the  square  of  a+6?  Of  a-j-J+c?  Of 
a-f-6-f-c-J-o??  According  to  what  law  is  the  square  of  any  polynomial 
formed?  By  reversing  this  law,  what  rule  have  we?  When  con- 
clude that  a  polynomial  is  not  a  perfect  square? 


176  RAYS  ALGEBRA,  FIRST  BOOK. 

197.  The  following  remarks  will  be  found  useful: 

1st.  iVo  binomial  can  be  a  perfect  square;  for  the  square 
of  a  monomial  is  a  monomial,  and  the  square  of  a  binomial 
is  a  trinomial. 

Thus,  a?-\-b2  is  not  a  perfect  square  ;  but  if  we  add  to  it 
2ab,  it  becomes  the  square  of  a-\-b.  If  we  subtract  from 
it  2ab,  it  becomes  the  square  of  a — b. 

2d.  In  order  that  a  trinomial  may  be  a  perfect  square, 
the  extreme  terms  must  be  perfect  squares,  and  the  middle 
term  twice  the  product  of  the  square  roots  of  the  extreme 
terms.  Hence,  to  find  the  square  root  of  a  trinomial  when 
it  is  a  perfect  square, 

Extract  the  square  roots  of  the  two  extreme  terms,  and  unite 
them  by  the  sign  plus  or  minus,  according  as  the  second  term 
is  plus  or  minus. 

Thus,  4a2 — 12ac-}-9c2  is  a  perfect  square ;  since  j/4a2— 2a, 
y9d^=3c,  and  +2aX— 3cX2=— 12ac. 

But  9x2-\-12xy-\-16y2,  is  not  a  perfect  square;  since 
j/^—Saj,  i/i6tf==&y,  and  SxX^X^—24:xy,  which  is 
not  equal  to  the  middle  term  12xy. 

GENERAL    REVIEW. 

What  is  meant  by  generalization?  Explain  by  an  example.  Rule 
for  finding  two  quantities  when  their  sum  and  difference  are  given. 
Rule  for  fellowship  without  time.  With  time.  What  is  a  negative 
solution,  and  what  does  it  imply? 

Explain  the  Problem  of  the  Couriers.  How  many  cases  may  be 
supposed?  Explain  each  case.  When  is  an  equation  independent? 
When  dependent?  When  redundant?  When  is  a  problem  inde- 
terminate? When  impossible?  Prove  that  a  simple  equation  has 
but  one  root. 

Rule  for  raising  a  monomial  to  any  power.  Rule  for  the  signs  in 
the  involution  of  monomials.  Why?  Rule  for  a  polynomial.  A 
fraction.  In  Newton's  theorem,  show  what  is  proved  in  regard  to 
number  of  terms.     Signs.     Exponents.     Coefficients. 

Why  can  no  binomial  be  a  perfect  square?  Example.  What  is 
necessary,  in  order  that  a  trinomial  may  be  a  perfect  square?  How 
may  its  square  root  be  found?     Example. 


RADICALS  OF  THE  SECOND  DEGREE.  177 

RADICALS 

OF    THE    SECOND    DEGREE. 

108.  From  Rule,  Art.  195,  it  is  evident  that  when  a 
monomial  is  a  perfect  square,  its  numeral  coefficient  is  a  per- 
fect square,  and  the  exponent  of  each  letter  is  exactly  divis- 
ible by  2. 

Thus,  4a2  is  a  perfect  square,  while  5a3  is  not. 

When  the  exact  division  of  the  exponent  can  not  be 
performed,  it  may  be  indicated,  by  writing  the  divisor 
under  it,  in  the  form  of  a   fraction.     Thus,  |/a3  may  be 

3 

written  a2. 

Since    a=a},    the   square   root  of  a  may   be  expressed 
j 
thus,  az.     Hence,  the  fractional  exponent,  ±,  is  used  to  in- 
dicate the  extraction  of  the  square  root. 


Thus,  }/dl-\-'lax^rxl  and  (a2-{-2ax-{-x2)'2,  also  y 4  and 

4~,  indicate  the  same   operation  ;  the  radical  sign  |/,  and 
the  fractional  exponent  i,  being  regarded  as  equivalent. 

Radicals  of  the  Second  Degree  are  quantities  affected 
by  a  square  root  sign  whose  root  can  not  be  exactly  found  ; 
as,  |/a,  -|/2,    a^/b,   and   5|/3;  or,  as  otherwise  written, 

a*,  23,  ab*,  and  5(3)2. 

Radicals  are  also  called  irrational  quantities,  or  surds. 

The  Coefficient  of  a  radical  is  the  quantity  which  stands 
before  the  radical  sign.  Thus,  in  the  expressions  aj/6, 
and  3|/5,  a  and  3  are  coefficients. 

Similar  Radicals  are  those  which  have  the  same  quan- 
tity under  the  radical  sign. 

Thus,  3j/2  and  7j/2  are  similar  radicals ;  so,  also,  are 
6j/a  and  2cy/a. 


178  RAY'S  ALGEBRA,  FIRST  BOOK. 

Radicals  that  are  not  similar,  may  frequently  become  so 
by  simplification.     This  gives  rise  to 


REDUCTION    OF    RADICALS. 

199.  Reduction  of  radicals  of  the  second  degree  con- 
sists in  changing  the  form  of  the  quantities  without  altering 
their  value.     It  is  founded  on  the  following  principle  : 

The  square  root  of  the  product  of  two  or  more  factors  is 
equal  to  the  product  of  the  square  roots  of  those  factors. 

That  is,  1/a6=1/aXy'6,   and  ^36=/ 9X/4;  for  -^36=^6, 

and  v/9Xv/ 4=3x2=6. 

Any  radical  of  the  second  degree  can,  on  this  principle, 
be  reduced  to  a  simpler  form,  when  it  can  be  separated  into 
factors,  one  of  which  is  a  perfect  square. 

Thus,  y'lT=1/4x3=1/4><  1/3=2/3" 

y/a3b=l/a2xoio=-]/a2X\/cib~=a-l/ab 
1/27a3c4=1/9a2c4x3a=1/9a2c4Xi/3a=3«c2v/3^    Hence, 


TO    REDUCE    A   RADICAL    OF    THE    SECOND   DEGREE   TO    ITS 
SIMPLEST  FORM, 

Rule. — 1.  Separate  the  quantity  into  two  parts,  one  of 
which  shall  contain  all  the  factors  that  are  perfect  squares. 

2.  Extract  the  square  root  of  the  part  that  is  a  perfect 
square,  and  prefix  it  as  a  coefficient  to  the  other  part  placed 
under  the  radical  sign. 

To  determine  whether  any  quantity  contains  a  numeral  factor 
that  is  a  perfect  square,  ascertain  if  it  is  divisible  by  either  of  the 
perfect  squares,  4,  9,  16,  25,  36,  49,  64,  81,  etc. 

Review. — 198.  When  is  a  monomial  a  perfect  square?  How  may 
the  square  root  of  a  quantity  be  expressed  without  the  radical  sign? 

198.  What  are  radicals  of  the  second  degree  ?  What  is  the 
coefficient  of  a  radical  ?     What  are  similar  radicals  ? 


RADICALS  OF  THE  SECOND  DEGREE.  179 

Reduce  each  of  the  following  radicals  to  its  simplest  form: 

i.  -|/8^: 


2.  !/12a3. 


Ans.  2aj/2. 
Ans.  2ay  3a. 

3.  l/20a*bs<*.  A.2abcy/ba~bc'. 

4.  V2TaV.    A.  12aCl/3a~c~. 

5.  7j/28aV".   A.  14a2c1/17^ 


6.1/32a66V.  A.  4a35cy2. 
7. 1/44a563c.  A.2a2Vll^". 

8.  l/48a866c4.  A.  4a*63c y  3~. 

9.  j/75aW!  A.  babcy'Sa'fc. 
10.1/243a3£2c.   A.9«y3a7. 


In    a   similar  manner,   polynomials   may   sometimes   be 
simplified. 


Thus.  !/  (2a3— 4a26+2a62)=1/  (a2— 2a&+&2)2a=(a— 6)j/2a. 

To  reduce  a  fractional  radical,  multiply  both  terms  by 
any  quantity  that  will  render  the  denominator  a  perfect 
square,  and  then  separate  the  fraction  into  two  factors,  as 
before  explained. 

11.  Reduce  j/j  to  its  simplest  form. 

l/F=l/IXf=l/l=V/4X^=l/i Xi/B=Ji/ 6,  Ans. 
Reduce  the  following  to  their  simplest  forms : 


12. 

i/i- 

Ans.  Ji/15. 

15.  9|/||*.       Ans.  fy3. 

13. 

VI 

Ans.  ly  14. 

16.  5T/T9o.    Ans.  iv710- 

14. 

vw 

Ans.  fv/3. 

17.  lOi/A-      Ans.  ^6. 

_ 

Since  a=i/a2,  3=1/9,  and  2l/3=1/4Xv/3=-|/4X3 
— -j/12,  it  is  obvious  that  any  quantity  may  be  reduced  to 
the  form  of  a  radical  of  the  second  degree,  by  squaring  it> 
and  placing  it  under  the  radical  sign. 

18.  Reduce  5  to  the  form  of  a  radical  of  the  second 
degree.  Ans.  -j/25. 

19.  Reduce  2a  to  the  form  of  a  radical  of  the  second 
degree.  Ans.  |/4a2. 


180  RAY'S  ALGEBRA.  FIRST  BOOK. 

20.  Express  3j/5,  entirely  under  the  radical. 

Ans.  j/45. 

21.  Pass  the  coefficient  of  Scy/2c,  under  the  radical. 

Ans.  j/lfc* 

ADDITION    OF    RADICALS. 

200.— 1.  What  is  the  sum  of  3^/2  and  5^2^ 

It  is  evident  that  3  times  and  5  times  any  certain  quan- 
tity must  make  8  times  that  quantity;  therefore, 

fy2+hy2=%l/2. 

2.  What  is  the  sum  of  2V//3~and  5j/T? 

Since  dissimilar  quantities  can  not  be  added,  we  can  only 
find  the  sum  of  these  expressions  by  placing  the  sign  of 
addition  between  them  ;  thus:  2|/3+5]/7. 

Sometimes  radicals  become  similar  after  being  reduced,  and 
may  then  be  added;  thus:  1/2-}-1/'8=1/2  +  21/2=3T/2. 
Hence, 

TO    ADD    RADICALS   OF   THE    SECOND   DEGREE, 

Rule. — 1.   Reduce  the  radicals  to  their  simplest  form. 

2.  If  the  radicals  are  similar,  add  the  coefficients,  and 
annex  the  common  radicals. 

3.  If  they  are  not  similar,  connect  them  by  their  proper 
signs. 

Find  the  sum  of  the  radicals  in  the  following  examples : 

3.  y^andj/TS". 


y/  8=1/4x2=21/2 
1/T8'=1/9x2=31/g 

Adding,  we  have  *>i/2j  Ans. 


RADICALS  OF  THE  SECOND  DEGREE.  181 

4.  yVl  and  j/27 Ans.  5/3. 

5.  /25  and  /80 Ans.  6/5T 

6.  /40,  /90,  and  /250 Ans.  lOylO. 

1.  /28^T2  and  /112a2&2 Ans.  6a6|/^T 

8.  y~{  and  1/^. 

Adding,  we  have  t8^/^'    Ans. 

9.  2/|  and  3/12 Ans.  7/3. 

10.  J-|/S  and  f/27 Ans.  /IF. 

11.  |/48o2c«a;  and  /12&2x.    .     .    Ans.  (4ac+2&)/3^. 

12.  Find  the  sum  of  j/(2a*— 4a*c+2ac*)  and 

/(2a3-f4a2c+2ac2).  Ans.  2a/2a. 

13.  Find  the  sum  of  y/a-\~x-\-y/ax2-\-x3-\-y/(a-\-xy. 

Ans.  (l-|-a+2a;)i/a-fa;. 

SUBTRACTION    OF    RADICALS. 
201.— 1.   Take  3/2  from  5/2. 

It  is  evident  that  5  times  any  quantity  minus  3  times  the 
quantity,  will  be  equal  to  2  times  the  quantity;  therefore, 

5/2-3/2=2/2. 
In  the  same  manner,     /8 — /2=2/2 — /2=r/2. 

Review. — 199.  In  what  does  reduction  of  radicals  of  the  second 
degree  consist?  On  what  principle  is  it  founded?  Illustrate  this 
principle. 

199.  Rule  for  the  reduction  of  a  radical?  How  determine  if  any- 
numerical  quantity  contains  a  factor  that  is  a  perfect  square?  How 
reduce  a  fractional  radical?     200.  Rule  for  addition  of  radicals. 


182  RAY'S  ALGEBRA,  FIRST  BOOK. 

If  the  radicals  are  dissimilar,  their  difference  can  only 
be  indicated.  Thus,  to  take  3/a  from  5/6,  write 
5]/6 — 3/a.     Hence, 


TO    SUBTRACT   RADICALS   OF   THE    SECOND   DEGREE, 

Rule. — 1.  Reduce  the  radicals  to  their  simplest  form. 

2.  If  the  radicals  are  similar,  find  the  difference  of  their 
coefficients,  and  annex  the  common  radical. 

3.  If  not  similar,  indicate  their  difference  by  the  proper 
sign. 

2.  ysM-yS. 

i/gZ=i/lgx2=V2 

1/""S=1/Tx2=2i/2 

Subtracting,  we  have         2/2,  Ans. 

3.  /45a2— /3a» Ans.  2a|/5? 

4.  y'bU—y'W. Ans.  2y'U. 

5.  /276V— i/ffi??. Ans.  bc-i/Sbc. 

6.  |/i5«5V— -|/25aW Ans.  2&c/a&7 

7.  5a/27— 3a/fg Ans.  8a|/5! 

8.  2/7—  3/T Ans.J). 

9.  yf-Vff- Ans.  T'H/30. 

10.  3/1— /2 Ans.  1/2. 

11.  /4a2#— a/a3'. Ans.  (2a—ax)y/x. 

12.  /3m2a;-f-6mnx-J-3?i2a; — /3m2cc — 6m7ix-|-3ri2a;. 

Ans.  2?i/3cc. 

MULTIPLICATION    OF    RADICALS. 

202.  Since  /a6=/aX/6,  therefore,  /a  X /&=/«&• 
See  Art.  199. 

Also,  a/lXc/d=aXcX/&X/^= ac/irf.     Hence, 


RADICALS  OF  THE  SECOND  DEGREE.  183 


TO    MULTIPLY   RADICALS    OF    THE    SECOND   DEGREE, 

Rule. — Multiply  the  coefficients  together  for  a  new  coefficient, 
and  the  quantities  under  the  radical  sign  for  a  new  radical. 

1.  Find  the  product  of  j/6  and  j/8. 

1/6Xi/8=1/48=v/16xB=41/3,  Ans. 

2.  Find  the  product  of  2^/TI  and  3j/2. 

2v/Mx31/^=6v/28=6v/4x7=6x21/7=12v/Y,Ans. 

3.  Find  the  product  of  j/8  and  j/27    .     .     .  Ans.  4. 

4.  Find  the  product  of  2-j/a  and  3-j/a.      .      Ans.  6a. 

5.  Find  the  product  of  ^727  and  j/3T  .     .     .  Ans.  9. 

6.  Find  the  product  of  3^2  and  2j/8.       Ans.  6j/6. 

7.  Find  the  product  of  21/15  and  3^35. 

Ans.  30^/21. 

8.  Find  the  product  of  -j/a3fc5c  and  y/abc.   Ans.  a263c. 

9.  Find  the  product  of  j/j  and  -j/f.         Ans.  TV|/5> 

10.  Find  the  product  of  2-*/^  and  3-*/'-^.   Ans.  -^-i/2. 

When   two   polynomials   contain    radicals,  they  may  be 
multiplied  as  in  multiplication  of  polynomials,  Art.  72. 

11.  Find  the  product  of  2-f-|/2~and  2—^/2.      Ans.  2. 

12.  Find  the  product  of  j/^+2~by  -y/x—2. 

Ans.  -j/x2 — 4. 


13.  Find  the  product  of  j/a-\-x  by  ^/a-\-x.  Ans.  a-j-a. 

14.  Find  the  product  of  j/x-\-2  by  j/a-f  3. 


Ans.  |/£C2-)-5a:-j-6. 

Review. — 201.  Rule  for  the  subtraction  of  radicals.     202.  For  the 
multiplication  of  radicals.     Prove  it. 


184  RAYS  ALGEBRA,  FIRST  BOOK. 

Perform  the  operations  indicated  in  the  following: 

15.  (C]/a-{-dl/b)X(cl/~a—dl/0).  .     .       Ans.  c2a— d2b. 

16.  (7  +  2/6)  X  (9— 5;/ 6) Ans.  3—1  7/6\ 

17.  (j/a-j-ic-f-j/a — x)(i/a-\-x — /a — x).  Ans.  2x. 

18.  (V— a:|/2  +  l)^2-fx,/24-l).   .     .     .     Ans.  x*-{- 1. 

DIVISION    OF    RADICALS. 
203.  Since  division  is  the  reverse  of  multiplication,  and 
since   y/aXy/b=i/ab,  therefore,  /ar6-i-/«=-*  / — =/&• 

Also,    since,    2/3x3/15=6/45  ;    therefore,  6/45 
--2/3=3/15.     Hence, 

TO   DIVIDE  RADICALS    OF    THE    SECOND   DEGREE, 

Rule. — Divide  the  coefficient  of  the  dividend  by  the  co- 
efficient of  the  divisor  for  a  new  coefficient,  and  the  radical 
of  the  dividend  by  the  radical  of  the  divisor  for  a  new 
radical. 

1.  Divide  8/72  by  2/6. 

8/72         _         _  

2-^=fl/V=Vl2=V4x3=8v/3,  Ans. 

2.  Divide  /54  by  j/&. Ans.  3. 

3.  Divide  6/54  by  3/27.      ....      Ans.  2/2. 

4.  Divide  /160  by  /87 Ans.  2/5. 

5.  Divide  15/378  by  5/6.    ....     Ans.  9/7. 

6.  Divide  abj/a3b3  by  i/a6 Ans.  a2b. 

7.  Divide  ^  by  yji  .     .     Ans.  ^  or  gj|/5W 

8.  Divide  /I  by  /-[. Ans.  J/6. 

Review.— 203.  Rule  for  the  division  of  radicals.     Prove  it. 


RADICALS  OF  THE  SECOND  DEGREE.  135 

9.  Divide  f^/18  by  ±y2 Ans.  4. 

10.  Divide  §yj  by  J^/J Ans.  fj/5. 

11.  Divide  i/i  by  T/2+31/i Ans.  T'G. 

20 1.  To  reduce  a  fraction  whose  denominator  contains 
a  radical  to  an  equivalent  fraction  having  a  rational  denom- 
inator. 

Since  y/aX|/«=r|/«2=«,  we  conclude  that  when  the 
square  root  of  a  quantity  is  multiplied  by  itself  or  squared, 
the  radical  sign  is  thrown  off. 

Thus,  y/2Xi/2=2,  and  v/^f6Xv/«T^=a+&- 
When   the  fraction   is   of  the  form  — — ,  if  we  multiply 
both  terms  by  -y/b,  the  denominator  will  become  rational. 

Thus,    -gL=  «xy6^«j/V 

Since  the  sum  of  two  quantities,  multiplied  by  their  dif- 
ference, is  equal  to  the  difference  of  their  squares  ;  if  the 
denominator  of  the  fraction  is  of  the  form  6-f-j/c,  and  we 
multiply  both  terms  by  b — j/c,  it  will  be  made  rational, 
since  it  will  be  b2 — c. 

Thus,     Xx^=^ 

6+jA  b—yc     b   c 

For  the  same  reason,  if  the  denominator  is  b — -j/c,  the 
multiplier  will  be  6+|/c.  If  it  is  j/&-f-]/c,  or  yb — -j/c, 
the  multiplier  will  be  -y/b — |/c,  or  \/b-\-^/c. 

These  different  forms  may  be  embraced  in  the  following 

General  Rule. — If  the  denominator  is  a  monomial,  mul- 
tiply both  terms  by  it;  but  if  it  is  a  binomial,  multiply  both 
terms  by  the  denominator,  with  the  sign  of  the  second  term 
changed. 

1st  Bk.  16 


186  RAY'S  ALGEBRA,  FIRST  BOOK. 

Reduce   the   following  fractions   to  equivalent  fractions, 
having  rational  denominators  : 

L  ^>l Ads.  Y=4y2. 

9    I72  a         l/6      .    /n 

8-  ^7g •  Ans.  T'T(6+j/F). 


Remark. — The  object  of  the  above  is  to  diminish  the  amount  of 

calculation  in  obtaining  the  numerical  value  of  a  fractional  radical. 

Thus,  suppose  it  is  required  to  obtain  the  numerical  value  of  the 

i/2 
fraction  — ■  in  example  2  above,  true  to  six  places  of  decimals. 

Here,  we  may  first  extract  the  square  root  of  2  and  of  3  to  seven 
places  of  decimals,  and  then  divide  the  first  result  by  the  second. 
This  operation  is  very  tedious.  If  we  render  the  denominator 
rational,  the  calculation  merely  consists  in  finding  the  square  root 
of  6,  and  then  dividing  by  3. 

Q 

5.  Find  the  numerical  value  of  -— .  Ans.  1.341 640 7-j-. 

1/* 

6.  Of      y2  _ Ans.  2.805883+. 


l/5-i/3 


SIMPLE    EQUATIONS    CONTAINING   RADICALS    OF 
THE    SECOND    DEGREE. 

205.  In  the  solution  of  questions  involving  radicals, 
much  will  depend  on  the  judgment  and  practice  of  the 
pupil,  as  almost  every  question  can  be  solved  in  several 
ways. 

The  following  directions  will  frequently  be  found  useful : 


SIMPLE  EQUATIONS  CONTAINING  RADICALS.        187 

1st.  When  the  equation  contains  one  radical  expression, 
transpose  it  to  one  side  of  the  equation,  and  the  rational 
terms  to  the  other,  then  involve  both  sides.     Thus, 

If  we  have  the  equation  \/(x — 1) — 1=2,  to  find  x. 

Transposing,  y  (X — 1)=  3 

Squaring,  X — 1  =  9 

X  =10. 

2d.  When  more  than  one  expression  is  under  the  radi- 
cal sign,  the  operation  must  be  repeated. 

Thus,  a+x=</(a?-\-Xi/c?+x*),  to  find  x. 
Squaring,  a2+2ax+x2=a2-\-xy'  c2+x2. 
Reducing  and  dividing  by  x,  2a-{-X=y,o2-\~x2. 
Squaring,  4a2 -f  4aX-\- x2=C2-\- x2\  whence,  x— — j . 

3d.  When  there  are  two  radical  expressions,  place  one 
of  them  alone  on  one  side,  before  squaring. 

Thus,  y'  (x— 5)— 3=4—  t/  (a;— 12),  to  find  x. 
Transposing,  |/  (x— 5)=7—  \/Jx— 12). 
Squaring,  x— 5=49— 14^ (x— 12) +x— 12. 
Reducing  and  transposing,  14-j/(iC — 12)=42. 
Dividing,  }/{x—\2)=Z. 
Squaring,  X — 12=9,  from  which  #=21. 

1.  y^+B-f 3=7 Ans.  z=13. 

2.  x -{-|/a2-f  11=11 Ans.  x=h. 

3.  1/(6+l/a=I)=3 Ans.  *=10. 

4.  i/x(a-\-x)=a — x Ans.  &=g. 

5.  y/x—2=l/x~^S Ans.  x=9. 

6.  jr-j-j/a:2— 7=7 Ans.  x=4. 

7.  1/a;+7=6— i/a;— 5 Ans.  cc=9. 

_  _  25a 

8.  ^/z— a=l/x— lySa Ans.  x=-y,  . 


188  RAYS  ALGEBRA,  FIRST  BOOK. 

9.  1/x-]-22q—1/x— 424— 11=0.     .    Ans.  a^lOOO. 

10.  x+y'Zax+tf—a Ans.  x=±a. 

_  -         -  5a 

11.  j/a-j-a— yx — a=-j/a Ans.  x=~^. 

12.  1/*+12=2+1/^T Ans.  a:=4. 

13.  \/S-]-x=:2y/l-\-x — j/sc Ans.  #=*. 

__         12  

14.  ^bx+^===l/lx+Q Ans.  »==§. 

ir       _  237— 1  Ox 

15.  1/^—4=— 7= Ans.  rc=23. 

4-fj/a; 

16.  ^/a;2+l/4^2-f£c+v/9x2+12x=l-fa;.       Ans.  «==J. 

17.  «*/a-j-j/aa;=j/a — %/a — j/aa;.  .     .     .     Ans.  a=|a. 

_       _  _        b(a+hy 

18.  b(i/x-\-y/b)=a(y/x— j/6).       .    Ans.  «=sv|Jl_^\r. 

19.  y/x-\-y/ax=a — 1 Ans.  aj=(j/u — l)2. 

GENERAL    REVIEW. 

Define  power.  Root.  Exponent.  Index.  Coefficient.  Factor. 
Term.  Square.  Square  root.  Cube.  Cube  root.  Surd.  Radical 
of  the  second  degree.  Rule  for  extracting  the  square  root  of  whole 
numbers.     Of  common  fractions.     Of  decimals. 

What  is  a  perfect  square?  An  imperfect  square?  Prove  that  the 
square  root  of  an  imperfect  square  can  not  be  a  fraction.  Rule  for 
extracting  the  square  root  of  an  algebraic  monomial.  Of  a  poly- 
nomial.    Prove  that  no  binomial  can  be  a  perfect  square. 

To  what  is  the  square  of  a  radical  of  the  second  degree  equal  ? 
How  reduce  an  integral  radical  to  its  simplest  form?  A  fractional 
radical.  Rule  for  addition  of  radicals.  Subtraction.  Multiplica- 
tion.    Division.     How  make  a  radical  denominator  rational? 

Define  elimination.  How  many  methods  of  elimination?  Define 
each.  Rule  for  each.  How  state  a  problem  containing  two  unknown 
quantities  ?  Containing  three  or  more?  When  is  the  first  method 
of  elimination  preferred?  The  second?  The  third?  What  two 
parts  in  the  solution  of  a  problem  ? 

When  the  denominator  of  a  fraction  contains  a  radical  of  the 
second  degree,  how  may  it  be  rendered  rational?  On  what  principle 
is  this  rule  founded? 


QUADRATIC  EQUATIONS.  189 

VII.    QUADRATIC    EQUATIONS. 

DEFINITIONS  AND  ELEMENTARY  PRINCIPLES. 

206.  A  Quadratic  Equation  is  an  equation  of  the 
second  degree  in  which  the  highest  power  of  the  unknown 
quantity  is  a  square;  as,  x2  =  9,  and  bx2  -j-  Sx  =  26. 

An  equation  containing  two  or  more  unknown  quantities 
is  of  the  second  degree,  when  the  sum  of  the  exponents 
of  the  unknown  quantities  in  any  term  is  2 ;  as,  xy—6} 
x2-\-xy=&,  and  xy-\-x-\-y=W. 

207.  Quadratic  equations  are  of  two  kinds,  pure  and 

affected. 

A  Pure  Quadratic  Equation  is  one  that  contains  the 
second  power  only  of  the  unknown  quantity,  and  known 
terms ;  as,  x2=9,  and  Sx2 — 5cc2=12. 

A  pure  quadratic  equation  is  also  called  an  incomplete 
equation  of  the  second  degree. 

An  Affected  Quadratic  Equation  is  one  that  contains 
both  the  first  and  second  powers  of  the  unknown  quantity, 
and  known  terms ;  as,  3x2-{-4x=20,  and  ax2 — bx2-\-dx 
—ex=f—g. 

An  affected  quadratic  equation  is  also  called  a  complete 
equation  of  tJu  second 


SOS.  Every  quadratic  equation  may  be  reduced  to  one 
of  the  forms  ax2=6,  or  ax2-\-bx=c. 

Review. — 206.  What  is  a  quadratic  equation?  Examples.  If  an 
equation  contains  two  unknown  quantities,  when  is  it  of  the  second 
decree?     Examples. 

207.  How  many  kinds  of  quadratic  equations?  What  is  a  pure 
quadratic?  Examples.  By  what  other  name  called?  What  is  an 
affected  quadratic?     Examples.     By  what  other  name  called? 


190  RAY'S  ALGEBRA,  FIRST  BOOK. 

In  a  pure  quadratic  equation,  all  the  terms  containing 
x2  may  be  collected  together,  and  its  form  becomes  ax2=b, 
or  ax2 — b=Q. 

An  affected  quadratic  equation  may  be  similarly  reduced ; 
for  all  the  terms  containing  xl  may  be  reduced  to  one  term, 
as  ax2 ;  and  those  containing  cc,  to  one,  as  bx ;  and  the 
known  terms  to  one,  as  c ;  then,  the  equation  is 

ax2-\-bx=c. 

PURE    QUADRATICS. 

2O0. — 1.  Let  it  be  required  to  find  the  value  of  x  in 
the  equation 

tf2— 16=0. 

Transposing,  £2=16. 

Extracting  the  square  root  of  both  members, 

x  =±4;  that  is,  £=-j-4,  or  — 4. 
Verification,     (+4)2— 16=16— 16=0. 
Or,     (—4)2—16=16—16=0. 

2.  Let  it  be   required   to  find  the  value    of  x   in    the 

equation 

5a;2-f-4=49. 
Transposing,  5#2=45. 

Dividing,  x2=  9. 

Extracting  the  square  root  of  both  sides, 

a:=dfc3. 

3.  Let  it  be  required  to  find  the  value  of  x  in  the 
equation 

2z2    3z2 

Clearing  of  fractions,  &c2-|-9:r2=68. 

Reducing,  17z2=68. 

Dividing,  X2=  4. 

Extracting  the  square  root,  X  =±2. 


QUADRATIC  EQUATIONS.  191 

4.  Given  ax2-\- b=cx2-\- d,  to  find  the  value  of  x. 

ax2 — cx2=d—b. 

Or,  (a— c)x2=d— b. 

o     d—b 


a — c 


-*£ 


6 

! — C* 


Hence, 


TO    SOLVE    A    PURE    QUADRATIC    EQUATION, 

Rule. — Reduce  the  equation  to  the  form  ax2=b.  Divide 
both  sides  by  the  coefficient  of  x2,  and  extract  the  square  root 
of  both  members. 

210.  If  we  take  the  equation  ax2—b,  we  have 

-     b 


x-=-. 
a 


x=±:^--  that  is,  *=-f^/-,  and  «=— yj^ 

If  we  assume  — — m2;  then,  x2=m2. 
a 

By  transposing,  X2 — m2=0. 

By  separating  into  factors,   (x-\-m)(x — m)=0. 

Now,  when  the  product  of  two  factors  is  0,  one  of  the  factors  must 
be  0.  Hence,  this  equation  can  be  satisfied  in  two  ways,  and  in  two 
only ;  that  is,  by  making  either  of  the  factors  equal  to  0. 

By  making  the  second  factor  equal  to  0,  we  have 
x — m=0,  or  x=i-\-m. 

I3y  making  the  first  factor  equal  to  0,  we  have 
#-f-m=0,  or  x= — m. 

Since  the  equation  (x-\-m)(x — m)=0,  can  be  satisfied  only  in  these 
two  ways,  it  follows  that  the  values  of  x  obtained  from  these  condi- 
tions are  the  only  values  of  the  unknown  quantity.     Hence, 

Review. — 208.  To  what  two  forms  may  every  quadratic  equation 
be  reduced?  Why?  209.  Rule  for  the  solution  of  a  pure  quadratic 
equation. 


192  RAY'S  ALGEBRA,  FIRST  BOOK. 

1.  Every  pure  quadratic  equation  has  two  roots,  and  only 
two. 

2.  These  roots  are  equal,  but  have  contrary  signs. 

Find  the  roots  of  the  equation,  or  the  values  of  x,  in 
each  of  the  following  examples : 

1.  a2— 8=28 Ans.  x=±zQ. 

2.  3x2— 15=83+x2 Ans.  x=±7. 

3.  a2x2 — b2=0 Ans.  X JL- 

a' 


4x2 
—1= 

5rc2 


4.  ^-1=27+1 Ans.  a*==fc8. 


5.  ^-+12=^+37f Ans.  x=^J. 


ax2 — b=(a — b)x2~\-c Ans.  x — '-L-\ 


6-fc 

~b~- 

i_    x — a       a — 2x     x24-bx  .  —y- 

7. =— t-1 — r Ans.  x=±i/ab. 

a  x — a       x — a 


QUESTIONS   PRODUCING  PURE    QUADRATIC    EQUATIONS. 

211. — 1.  Find  a  number  whose  §  multiplied  by  its  §, 

equals  60. 

2.x    2*c     A.x^ 
Let  x=  the  number;  then,  -7r-X-?-=^r^=  60. 
3  /x  5        15 

4^=900. 

#2=225. 
x=  15. 

2.  What  number  is  that,  of  which  the  product  of  its  third 
and  fourth  parts  is  equal  to  108?  Ans.  36. 

3.  What  number  is  that  whose  square  diminished  by  16, 
is  equal  to  half  its  square  increased  by  16?  Ans.  8. 

Review. — 210.  Show  that  every  such  equation  has  two  roots,  and 
only  two.     That  they  are  equal,  but  have  contrary  signs. 


QUADRATIC  EQUATIONS.  193 

4.  What  number  is  that,  which  being  divided  by  9,  gives 
the  same  quotient  as  16  divided  by  the  number  ?  Ans.  12. 

5.  "What  two  numbers  are  to  each  other  as  3  to  5,  and 
the  difference  of  whose  squares  is  64  ? 

Let  3x=  the  less  number ;  then,  5a?=  the  greater. 
And  (5x)2—  (3z)2=64. 

Or,         25x2—9x2=16x2=64. 
X  =  2;  hence, 
3x  =  6,  and  5z=10. 

6.  What  two  numbers  are  to  each  other  as  3  to  4,  the 
difference  of  whose  squares  is  63  ?  Ans.  9  and  12. 

7.  The  breadth  of  a  lot  is  to  its  length  as  5  to  9,  and 
it  contains  1620  sq.  ft. ;  required  the  breadth  and  length. 

Ans.  Breadth  30,  length  54  ft. 

8.  Find  two  numbers  whose  sum  is  to  the  greater  as 
10  to  7,  and  whose  sum,  multiplied  by  the  less,  is  270. 

Ans.  21  and  9. 

Let  10:r=r  their  sum ;  then,  7X—  the  greater,  and  3x=  the  less. 

9.  What  two  numbers  are  those,  whose  difference  is  to 
the  greater  as  2  to  9,  and  the  difference  of  whose  squares 
is  128?  Ans.  18  and  14. 

10.  A  person  bought  a  piece  of  muslin  for  $3  and  24  cts., 
and  the  number  of  cts.  which  he  paid  for  a  yd  ,  was  to  the 
number  of  yd.  as  4  to  9  ;  how  many  yd.  did  he  buy,  and 
what  was  the  price  per  yd.  ? 

Ans.  27  yd.,  at  12  cts.  per  yd. 

11.  Find  two  numbers  in  the  ratio  of  *  to  §,  the  sum 
of  whose  squares  is  225.  Ans.  9  and  12. 

Reducing  |  and  §  to  a  common  denominator,  we  find  they  are 
as  3  to  4.     Then,  let  3a;  and  4x  represent  the  numbers. 

12.  Find  three  numbers  in  the  proportion  of  ^,  f ,  and 
|,  the  sum  of  whose  squares  is  724. 

Ans.  12,  16,  and  18. 

1st  Bk.  17* 


194  RAY'S  ALGEBRA,  FIRST  BOOK. 

AFFECTED    QUADRATIC    EQUATIONS. 

1.  Required  to  find  the  values  of  x  in  the  equation 

x2— 4x-j-4=l. 

It  is  evident,  from  Art.  197,  that  the  first  member  of  this  equation 
is  a  perfect  square.  By  extracting  the  square  root  of  both  members, 
we  have  x — 2=  =hl. 

Whence,  a=2±l=2+l=3,  or  2—1=1. 

Verification,     (3)2— 4x3+4=1 ;  that  is,  9—12+4=1. 

Also,  (1)2—4x1+4=1 ;  that  is,  1-  4+4=1. 

Hence,  x  has  two  values,  +3  and  +1,  either  of  which  verifies  the 
equation. 

2.  Required  to  find  the  value  of  x,  in  the  equation 

x2+6x=16. 

If  the  left  member  of  this  equation  were  a  perfect  square,  we 
might  find  the  value  of  #,  by  extracting  the  square  root,  as  in  the 
preceding  example. 

By  a  careful  examination  of  the  principle  stated  in  Art.  197,  we 
discover  that  the  first  member  will  become  a  perfect  square  if  9  be 
added  to  it. 

Adding  9  to  each  member, 

x2+6x+9=25. 

Extracting  the  square  root,  a:+3=dz5. 

Whence,  x=— 3zfc5=+2,  or  —8. 

Either  of  which  values  of  X  will  verify  the  equation. 

212.  Every  affected  quadratic  equation  is  of  the  form 

ax2-\-bx=c. 

If  we  divide  both  sides  by  a,  and  make  the  term  in  x2  positive, 
there  can  be  but  four  possible  forms,  according  as  the  signs  of  the 
other  two  terms  are  positive  or  negative,  viz. : 

x2+2px=q   (1) 
x2 — 2px=q  (2) 
x2-\-2px=— q  (3) 
x2— 2px= — q  (4) 
In  which  2  p  and  q  may  be  either  integral  or  fractional. 


QUADRATIC  EQUATIONS.  195 

We  will  now  explain  the  principle  by  which  the  first  member  of 
this  equation  may  always  be  made  a  perfect  square. 

The  square  of  a  binomial  is  equal  to  the  square  of  the  first  term, 
plus  twice  the  product  of  the  first  term  by  the  second^  plus  the 
square  of  the  second. 

If,  now,  we  consider  x2-{-2px  as  the  first  two  terms  of  the  square 
of  a  binomial,  x2  is  the  square  of  the  first  term  (x),  and  2px,  the 
double  product  of  the  first  term  by  the  second ;  therefore, 

If  we  divide  2px  by  2x,  the  quotient,  p  {half  the  coefficient  of  x), 
will  be  the  second  term  of  the  binomial,  and  its  square,  p2,  added  to 
the  first  member,  will  render  it  a  perfect  square.  To  preserve  the 
equality,  we  must  add  the  same  quantity  to  both  sides.     This  gives 

x2+2px+p2=q+p2 


Extracting  the  square  root,  X-\-  p=±v/q-^-p2 


Transposing,  x=—pzt:l/q-\-p2. 

It  is  obvious,  that  the  square  may  be  completed  in  each  of  the 
other  forms,  on  the  same  principle. 

Collecting  the  values  of  x  in  each,  we  have  the  following  table: 

(1.)    x2+2px=q.  x=—p±l/ q-\-p2. 

(2.)    x2—2px=q.  x=-{-pz±zl/  q-\-p2. 


(3.)    x2+2px=— q.  x=—p±zl/—q-\-p2. 

(4.)    x2— 2px=— q.  x=-\-p  rfc  j/ — q-\-p2.     Hence, 


TO    SOLVE   AN    AFFECTED    QUADRATIC   EQUATION, 

Rule. — 1.   Reduce   (he  equation,   by  clearing  of  fractions 
and  transposition  if  necessary,  to  the  form  ax2-\-bx=c. 

2.  Make  the  first  term  positive,  if  it  is  not  so  already. 

3.  Divide  each  side  of  the  equation  by  the  coefficient  of  x*. 

4.  Add  to  each  member  the  square  of  half  the  coefficient  ofx. 

5.  Extract  the  square  root  of  both  sides. 

6.  Transpose  the  known  term  to  the  second  member. 


196  RAY'S  ALGEBRA,  FIRST  BOOK. 

1.  Find  the  roots  of  the  equation  x2-\-Sx—SS. 

Completing  the  square  by  taking  half  the  coefficient  of  x,  squar- 
ing it,  and  adding  to  each  member,  we  have 

z2_}_8a;-{-16=33-}-16=49. 
Extracting  the  root,      x-\-  4=±7. 
Transposing,  x= — 4±7. 

Whence,  x=— 4-}-7=-}-3. 

And  x=— 4— 7=— 11. 

Verification.     (3)2-j-8(3)=33 ;  that  is,      9+24=33. 
Or,        (— ll)2_f-8(— 11)=33;  that  is,  121— 88=33. 

2.  Solve  the  equation  x2 — 6x=16. 

Completing  the  square,   X2 — 6z-|-9=16-{-9=25. 
Extracting  the  root,  x — 3=±5. 

Transposing,  x=-\-3dz5. 

Whence,  #=-f  3-j-5=-}-8. 

And  x=+3-5=—2. 

Both  of  which  will  be  found  to  verify  the  equation. 

3.  Solve  the  equation  x2-\-6x= — 5. 

Completing  the  square,  x2-\-6x-\-9=9 — 5=4. 
Extracting  the  root,  #-f-3=±2. 

Transposing,  x= — 3±L2. 

Whence,  x=— 3-|-2=— 1. 

And  x=—%—2=—5. 


4.  Find  the  values  of  x,  in  the  equation  x2 — 10:r= — 24. 

Completing  the  square,  x2 — 10a;-f-25=25 — 24=1. 
Extracting  the  root,  x — 5=zhl. 

Transposing,  a:=5±l. 

Whence,  z=5-fl=6. 

And  ic=5 — 1=4. 

The  preceding  examples  illustrate  the  four  different  forms,  when 
the  equation  is  already  reduced.  Generally,  however,  equations 
are  more  complicated,  and  require  to  be  reduced  before  completing 
the  square. 


QUADRATIC  EQUATIONS. 


197 


5.  Find  the  values  of  x,  in  the  equation  Sx — 5: 


7z+36 


Clearing  of  fractions, 

Transposing, 

Dividing, 

Completing  the  square, 

Extracting  the  root, 

Transposing, 

Whence, 


3x2—  5x=7x-\-36. 
3x2—12x=36. 
x2—  4z=12. 
x2—  4a:-|-4=16. 
x— 2==fc4. 

z=6,  or  —2. 


6.  Find  the  values  ofx,  in  the  equation 


12: 


;=52+ 


13c 


Clearing  of  fractions,  12a:2-f-5a:=260-|-13a;. 

Transposing  and  reducing,    lZx2 — &c=260. 
Dividing,  x2—2x=G£. 

Here,  the  coefficient  of  x  is  — |,  the  half  of  which  is  — 
square  of  this  is  1,  which  being  added  to  both  sides,  we  have 

L5. 


the 


Extracting  the  root, 
Whence, 


X=-\-5,  or  — 


Note  . — The  following  examples  illustrate  the  four  forms,  to  one 
of  which  every  complete  equation  of  the  second  degree  may  be  re- 
duced. 


7.  *2-j-8:z=20. 

8.  x2+16x=%0. 

9.  z2-{-3:r=28. 
10.  x2— 10a;=24. 


11. 


5x=6.  . 


12.  x'z+6x=—  8. 

13.  x*+1x=— 12 


.  Ans.  x=2,  or  — 10. 
.  Ans.  x=4,  or  — 20. 
.  Ans.  x=4,  or  — 7. 
.  Ans.  #=12,  or  — 2. 


0, 


— 1 


Ans.  x= — 2,  or  ■ — 4. 
Ans.  x= — 3,  or  — 4. 


Review. — 212.  To  what  general  form  may  every  affected  quadratic 
be  reduced?  What  are  the  four  forms  that  this  gives,  depending 
on  the  signs  of  2p  and  ql 

212.  Explain  the  principle,  by  means  of  which  the  first  member 
of  the  equation  z2-\-2px=q  may  be  made  a  perfect  square.  Rule 
for  the  solution  of  affected  quadratic  equations. 


198  RAY'S  ALGEBRA,  FIRST  BOOK. 

14.  x2 — 8x= — 15 Ans.  a;=5,  or  3. 

15.  x* — lbx= — 54 Ans.  x=9,  or  6. 

16.  3a;2— 2a;+123=256.     .     .     .   Ans.  x=1,  or  —  y>. 

17.  2a;2—  5a;=12 Ans.  a;=4,  or  .-j. 

18.  -j -5-=| Ans.  a;=4,  or  — f 

19.  a2— a;— 40=1 70.      .     .     .     Ans.  a;=15,  or  —14. 

20.^ ^o=4 Ans.  a;=24,  or -6. 

4      a— 2  ' 

21.  %x2—  >aj+j=8— fa;— aj*+^.  Ans.  a;=4,  or—  ff. 

22.  a;2+a;=30 Ans.  «=5,  or—  6. 

23.  g+?=4+j Ans.  ^2,  or  4. 

2  ■  a;    4  '  a 

24.  2a;2+92=31a; Ans.  a;=4,  or  11^. 

25.  —  x2-\-x=t% Ans.  x=%  or  f. 

26.  17a;2— 19x=30 Ans.  a;=2,  or—  }§. 

27.  4a5— 3a;2=6a;— 8 Ans.  as=f,  or  —2. 

28.  a;2— 4a;=— 1. 

Ans.  a;=2±v/3=3.732+,  or  .268—. 
OQ    4o5       2.x2     10a;       20  .  , 

y  — "3-=-3 y-  •    •    •    Ans-  x=—»,  <>*  ?• 

80-  -T8=9^XT Ans.  *=12,  or -2. 

a;+8     2a5+l 

24 

31.  a?H T=3a; — 4 Ans.  #=5,  or  — 2. 

x — 1 

QO    a;+3  ,     7a;       23  a  /.  i 

^'S&ili&T*    •     {     •  An,  a5=8,  or  13if. 
34..  2ax—x2=—2ab—b\      .     .  Ans.  a;=2a+5,  or  —b. 

35.  ar2+36z— 462=0.       .     .     .    Ans.  a:=+6,  or  —46. 

36.  x2 — ax — bx= — ab Ans.  #=+a,  or  +6. 

37.  2&s2+(a— 2b)x=a Ans.  x=l,  or  —  ~. 

38.  x1 — (a — V)x — a=0 Ans.  x—a,  or  — 1. 

39.  x2 — (a+6 — c)x=(a-\-b)c    .      Ans.  a;=a+&,  or  — c. 


QUADRATIC  EQUATIONS.  199 

213.  The  Hindoo  method  of  solving  quadratics. — 
When  an  equation  is  brought  to  the  form  ax2-\-bx=c,  it 
may  be  reduced  to  a  simple  equation,  without  dividing  by 
the  coefficient  of  x2 ;  thus  avoiding  fractions. 

If  we  multiply  both  sides  of  the  equation  ax2-{-bx=c,  by  a,  the 
coefficient  of  x2,  it  becomes  a2x2-\-abx=ac. 

Now,  we  may  regard  a2x2-\-abx,  or  a2x2-\-bax,  as  the  first  and 
second  terms  of  the  square  of  a  binomial,  a2x2  being  the  highest 
power,  ax  the  lower  power  with  a  coefficient  b. 

Completing  the  square  by  adding  the  square  of  ^  to  each  side,  the 

b2  b2  A  ■ 

equation  becomes  a2x2-\-abx-^—j-=ae-\—j-. 

Now,  the  left  side  is  a  perfect  square ;  but  it  will  still  be  a  perfect 
square,  if  we  multiply  both  sides  by  4,  since  the  product  of  a  square 
number  by  a  square  number  is  always  a  square  number. 

Multiplying  by  4,  the  equation  is  cleared  of  fractions,  and  we 
have  4a2x2-\- 4abx-\-b2=4ac-\-b2. 

Extracting  the  square  root, 


2a^+6=d=v/4ac+62. 


Whence,  x= £ ^— . 

The  equation  Aa2x2-\-^abx-\-b2=kac-\-b2,  may  be  derived  directly 
from  the  equation  ax2-\- bx=c,  by  multiplying  both  sides  by  4a,  the 
coefficient  of  X2,  and  then  adding  to  each  member,  the  square  of  6, 
the  coefficient  of  the  first  power  of  x.     Hence, 


TO    SOLVE   AN   AFFECTED   QUADRATIC   EQUATION, 

Rule. — 1.   Reduce  the  equation   to  the  form  ax2-\-bx=c, 
and  multiply  both  sides  by  four  times  the  coefficient  of  x2. 

2.   Add  the  square  of  the  coefficient  of  x  to  each  side,  and 
extract  the  square  root. 

Review. — 213.  Explain    the  Hindoo    method  of  completing   the 
square. 


200  BAY'S  ALGEBRA,  FIRST  BOOK. 

1.  Given  Sx2 — &x=28,  to  find  the  values  of  x. 

Multiplying  both  sides  by  12,  which  is  4  times  the  coefficient  of  X2, 
36x2— 6(te=336. 

Adding  to  each  member  25,  the  square  of  5,  the  coefficient  of  x, 
36z2— 60z-f  25=361. 

Extracting  the  root,  6x —  5^=  ±19. 

6z=5±19z=24,  or  —14. 


x=-\-4,  or  — 


By  the  same  rule,  find  the  values  of  the  unknown  quan- 
tity in  each  of  the  following  examples: 

2.  2x2-f  5^=33 Ans.  x=3,  or  — -U-. 

3.  ox2+2x--=$$ Ans.  as==4,  or  —  232-. 

4.  Sx2—  x=70 Ans.  x=Z>,  or—-1/. 

5.  x2 —  z=42 Ans.  cc=7,  or    — 6. 

6.  |z2+^—  5=9j Ans.  x=6,  or  —  1±. 


PROBLEMS    PRODUCING    AFFECTED    QUADRATIC 
EQUATIONS. 

214. — 1.  "What  number  is  that,  whose  square  dimin- 
ished by  the  number  itself,  is  equal  to  20  ? 

Let  x=  the  number. 
Then,  x2— rr=20. 

Completing  the  square,  x2— x  -|_l— 20-f-  \=§£. 
Extracting  the  root,  x — i=±t. 

Whence,  #=-|-5,  or  — 4. 

The  negative  value  — 4,  will  answer  the  conditions  of  the  ques- 
tion in  an  arithmetical  sense,  if  the  question  be  changed,  thus : 
what  number  is  that,  whose  square  increased  by  the  number  itself, 
is  equal  to  20  ? 


QUADRATIC  EQUATIONS.  201 

2.  A  person  buys  several  oranges  for  60  cts. ;  had  he 
bought  3  more  for  the  same  sum,  each  orange  would  have 
cost  him  1  ct.  less ;  how  many  did  he  buy  ? 

Let  X—  the  number  bought. 

fiO 
Then,  — —  the  price  of  each. 

fin 

And  — j-q=  the  price  of  one,  had  he  bought  3  more  for  60  cts. 

X-J[-  o 

rm.       ,  60         60       t 

Therefore,  --^—^=1. 

Clearing  of  fractions,  and  reducing, 

x*-\-Sx=\W. 
Completing  the  square,  x2-\-^x-\-^^-\-\%Q='J ^ . 

Extracting  the  root,  x-\-^=±%?. 

Whence,  a=+12,  or  —15. 

Either  of  these  values  satisfies  the  equation  from  which  it  was 
derived ;  but  only  one  satisfies  the  conditions  of  the  question. 

Since  ;r105= — 4  and  -^-3= — 5;  and  since  buying  and  sell- 
ing are  opposite,  the  result,  — 15,  is  the  answer  to  the  following 
question : 

A  person  sells  several  oranges  for  60  cts.  Had  he  sold  3  less  for 
the  same  sum,  he  would  have  received  1  ct.  more  for  each;  how  many 
oranges  did  he  sell  ? 

Remark. — From  the  two  preceding  examples,  we  see  that  the 
positive  root  satisfies  both  the  conditions  of  the  question,  and  the 
equation  derived  from  it;  while  the  other  root  satisfies  the  equation 
only. 

The  negative  value  is  the  answer  to  a  question,  differing  from  the 
one  proposed,  in  this;  that  certain  quantities  which  were  additive, 
have  been  subtractive,  and  vice  versa. 

Sometimes,  however,  as  in  the  following  example,  both  values  of 
the  unknown  quantity  satisfy  the  conditions  of  the  question. 

3.  Find  a  number,  whose  square  increased  by  15,  shall 
be  8  times  the  number. 

Let  x=  the  number;  then,  £2-f-15=&e. 

Or,  x2— 8x=— 15. 

Whence,  x=5,  or  3. 

Either  of  which  fulfills  the  conditions  of  the  question. 


202  RAY'S  ALGEBRA,  FIRST  BOOK. 

When  a  problem  contains  two  unknown  quantities,  and  can  be 
solved  by  the  use  of  one  symbol,  the  two  values  generally  give  the 
values  of  both  unknown  quantities,  as  in  the  following : 

4.  Divide  24  into  two  such  parts,  that  their  product 
shall  be  95. 

Let  #=  one  of  the  parts ;  then,  24 — #=  the  other. 
And  #(24— #)=95. 

Or,  #2— 24#=— 95. 

Whence,  #=19  and  5. 

And  24— #=5,  or  19. 

5.  Find  3  numbers,  such  that  the  product  of  the  first 
and  third  is  equal  to  the  square  of  the  second ;  the  sum 
of  the  first  and  second  is  10,  and  the  third  exceeds  the 
second  by  24. 

Let  #=  the  first ;   then,  10 — x=  the  second. 
And  10—  #+24=34— #=  the  third. 

Also,  (10— #)2=#(34— #). 

Or,  100— 20#+#2=34#— x2. 

From  which,  #=25,  or  2. 

When  #=25,  10— #=—15,  34— #=9,  and  the  numbers  are  25, 
—15,  and  9. 

When  #=2,  10— #=8,  34— #=32,  and  the  numbers  are  2,  8, 
and  32. 

Both  sets  of  values  satisfy  the  question  in  an  algebraic  sense; 
only  the  last  in  an   arithmetical  sense. 

The  first  will  satisfy  the  conditions  of  the  following: 
Find  three  numbers,  such  that  the  product  of  the  first  and  third 
is  equal  to  the  square  of  the  second-,  the  difference  of  the  first  and 
second  is  10;  and  the  sum  of  the  second  and  third  is  24. 


Remark. — In  the  following  examples,  it  is  required  to  find  only 
that  value  of  the  unknown  quantity  which  satisfies  the  conditions 
of  the  question  in  an  arithmetical  sense.  It  forms,  however,  a  good 
exercise  to  determine  the  negative  value,  and  modify  the  question, 
as  above. 


QUADRATIC  EQUATIONS.  2C3 

6.  Find  a  number,  such  that  if  its  square  be  diminished 
by  6  times  the  number  itself,  the  remainder  shall  be  7. 

Ans.  7. 

7.  Find  a  number,  such  that  twice  its  square,  plus  3  times 
the  number  itself,  shall  be  65.  Ans.  5. 

8.  Find  a  number,  such  that  if  its  square  be  diminished 
by  1,  and  §  of  the  remainder  be  taken,  the  result  shall  be 
equal  to  5  times  the  number  divided  by  2.  Ans.  4. 

9.  Find  two  numbers  whose  difference  is  8,  and  whose 
product  is  240.  Ans.  12  and  20. 

10.  A  person  bought  a  number  of  sheep  for  $80  ;  if 
he  had  bought  4  more  for  the  same  money,  he  would  have 
paid  $1  less  for  each  ;  how  many  did  he  buy?     Ans.  16. 

11.  There  are  two  numbers,  whose  difference  is  10,  and 
if  600  be  divided  by  each,  the  difference  of  the  quotients 
is  also  10  ;  what  are  the  numbers  ?        Ans.  20  and  30. 

12.  A  pedestrian,  having  to  walk  45  mi.,  finds  that  if 
he  increases  his  speed  J  mi.  an  hr.,  he  will  perform  his 
task  1\  hr.  sooner  than  if  he  walked  at  his  usual  rate  ; 
what  is  that  rate?  Ans.  4  mi.  per  hr. 

13.  Divide  the  number  14  into  two  parts,  the  sum  of 
whose  squares  shall  be  100.  Ans.  8  and  6. 

14.  In  an  orchard  of  204  trees,  there  are  5  more  trees 
in  a  row  than  there  are  rows;  required  the  number  of  rows, 
and  of  trees  in  a  row.     Ans.  12  rows,  17  trees  in  a  row. 

15.  A  and  B  start  at  the  same  time  to  travel  150  mi.  ; 
A  travels  3  mi.  an  hr.  faster  than  B,  and  finishes  his  jour- 
ney 8 1  hr.  before  him ;  at  what  rate  per  hr.  did  each 
travel?  Ans.  9  and  6  mi.  per  hr. 

16.  A  company  at  a  tavern  had  $1  and  75  cts.  to  pay; 
but  before  the  bill  was  paid  two  of  them  went  away,  when 
those  who  remained  had  each  10  cts.  more  to  pay;  how 
many  were  in  the  company  at  first?  Ans.  7. 


204  RAY'S  ALGEBRA,  FIRST  BOOK. 

17.  The  product  of  two  numbers  is  100,  and  if  1  be 
taken  from  the  greater  and  added  to  the  less,  the  product 
of  the  resulting  numbers  is  120;  what  are  the  numbers? 

Ans.  25  and  4. 

18.  If  4  be  subtracted  from  a  father's  age,  the  remain- 
der will  be  thrice  the  age  of  the  son  j  and  if  1  be  taken 
from  the  son's  age,  half  the  remainder  will  be  the  square 
root  of  the  father's  age.     Required  the  age  of  each. 

Ans.  49  and  15  yr. 

19.  A  young  lady  being  asked  her  age,  answered,  "  If 
you  add  the  square  root  of  my  age  to  |  of  my  age,  the 
sum  will  be  10."     Required  her  age.  Ans.  16  yr. 

20.  Bought  a  horse,  which  I  afterward  sold  for  $24,  thus 
losing  as  much  per  cent,  upon  the  price  of  my  purchase  as 
the  horse  cost;  what  did  I  pay  for  him? 

Ans.  $60  or 


PROPERTIES  OF  THE  ROOTS  OF  AN  AFFECTED 
QUADRATIC  EQUATION. 

215.  The  preceding  examples  show  that  in  a  quadratic 
equation,  the  unknown  quantity  has  two  values.  This  prin- 
ciple may  be  proved  directly,  as  follows : 

Take  the  general  form,  x2-\-2px=q,  in  which  2p  and  q  may  be 
either  both  positive  or  both  negative,  or  one  positive  and  the  other 
negative.     Completing  the  square, 

We  have  x2-\-2px+p2=q+p2. 

Assume  q-\- p2=m2 . 

That  is,  y/q-\-pY=m ; 

Then,  (x+p)2=m2. 

Transposing,  (x+p)2—m2=0. 

Resolving  into  factors,  (x+p-\-m)(x-\-p — m)=Q. 

Now,  this  equation  can  be  satisfied  in  two  ways,  and  in  only  two; 
that  is,  by  making  either  of  the  factors  equal  to  0,  Art.  210. 
If  we  make  the  second  factor  equal  to  zero, 
We  have  x-\-p — ra=0. 

Or,  by  transposing,  X=— p-}-m=— p-\- y/q+p?. 


QUADRATIC  EQUATIONS.  205 

If  we  make  the  first  factor  equal  to  zero, 

We  have  x-\-p-\-m=0. 

Or,  by  transposing,  X-—p—m=—p — •/q+p2.     Hence, 

1.  Every  quadratic  equation  has  two  roots,  and  only  two. 

2.  Every  affected  quadratic  equation,  reduced  to  the  form 
x2-\-2px=q  may  be  decomposed  into  two  binomial  factors, 
of  which  the  first  term  in  each  is  x,  and  the  second,  the  two 
roots  with  their  signs  changed. 

Thus,  the  two  roots  of  the  equation  x2 — 5x= — 6,  or  x2—* 5x-{- 6— 0, 
are  x=2  and  x=3;  hence,  x2 — bx-\-G=(x — 2){x — 3). 

From  this,  it  is  evident  that  the  direct  method  of  resolving  a  quad- 
ratic trinomial  into  its  factors,  is  to  place  it  equal  to  zero,  and  then 
find  the  roots  of  the  equation. 

In  this  manner,  let  the  learner  solve  the  questions  in  Art.  95. 

By  reversing  the  operation,  we  can  readily  form  an  equation, 
whose  roots  shall  have  any  given  values.     Thus, 

Let  it  be  required  to  form  an  equation  whose  roots  shall  be 
4  and  —6. 

We  must  have  x=     4  or  x — 4=  0. 

And  z=— 6  ora:+6=:  0. 

Hence,  (x— 4)(x+6)=:X2 +2x— 24=  0. 

Or,  x2+2x=24:. 

Which  is  an  equation  whose  roots  are  -(-4  and  — 6. 

1.  Find  an  equation  whose  roots  are  7  and  10. 

Ans.  x2— 11x=— 70. 

2.  Whose  roots  are  — 3  and  — 1.     Ans.  x2-\-4tx= — 3. 

3.  Whose  roots  are  -|-2  and  — 1.  Ans.  x2 — x=-2, 

216.  Resuming  the  equation  x2-\-2px=q. 


The  first       value  of  x  is        — p-\-\/q-\-p2. 
The  second  value  of  x  is        — p—  \/q-\-p2. 

Their  sum  is  — 2p,  which  is  the  coefficient 

of  x,  taken  with  a  contrary  sign.     Hence, 


206  RAY'S  ALGEBRA,  FIRST  BOOK. 

The  sum  of  the  roots  of  a  quadratic  equation  reduced  to  the 
form  x2-\-2px—q,  is  equal  to  the  coefficient  of  the  first  power 
of  x  taken  with  a  contrary  sign. 

If  we  take  the  product  of  the  roots,  we  have 


First  root,  =—PJrV(lJrP' 


Second  root,  =— p—  i/q-{-p2 


P2-PV<2+P2 

h2Vg+p2— (q+p2) 
p2  •   •   •   --{q~\-p2)=- q. 


But  — q  is  the  known  term  of  the  equation,  taken  with  a  contrary 
sign.     Hence, 


The  product  of  the  two  roots  of  a  quadratic  equation,  re- 
duced to  the  form  x2-{-2px=q,  is  equal  to  the  known  term 
taken  with  a  contrary  sign. 

Remark. — In  the  preceding  demonstrations,  we  have  regarded 
2p  and  q  as  positive;  the  same  course  of  reasoning  will  apply  in 
each  of  the  four  different  forms. 

til 7.  In  the  equation  x2-\-2px—q,  or  first  form,  the  two 
values  of  x  are 

— p+\/q+p2 


And  —p—yq+P2- 


The  value  of  the  part  y1 q+p2  must  be  a  quantity  greater  than  p, 
since  the  square  root  of  p2  alone  is  p.     Hence, 

The  first  root  is  equal  to  — p  plus  a  quantity  greater  thanp; 
therefore,  it  is  essentially  positive. 

The  second  root  being  equal  to  the  sum  of  two  negative  quan- 
tities, p,  and  a  quantity  greater  than  p,  is  essentially  negative. 

It  is  also  obvious,  that  the  second  or  negative  root  is  numerically 
greater  than  the  first,  or  positive  root.  See  problems  7,  8,  9, 
Art.  212. 


QUADRATIC  EQUATIONS.  207 

In   the   equation  x2 — 2px=q,   or   second  form,   the   two   values 
of  X  are 

+P+W+P2 


And  +P—  V^rP1- 

Reasoning  as  before,  we  find  that  the  first  root  is  essentially  posi- 
tive, and  greater  than  2p. 

The  second  root  is  equal  to  p,  minus  a  quantity  greater  than  p ; 
therefore,  it  is  essentially  negative. 

The  first,  or  positive  root,  is  evidently  greater  than  the  second,  or 
negative  root.     See  problems  10  and  11,  Art.  212. 

In   the   equation  x2-\-2px-= — q,  or  third  form,  the  two  values 
of  x  are 


— P+V—  <2+P2 
And  — p— ]/— q-\-p2. 


Here,  the  value  of  -j/ — q-\-p2,  is  less  than  p;  hence,  the  first  root 
is  — p,  plus  a  quantity  less  than  p;  therefore,  it  is  essentially 
negative. 

It  is  plain  that  the  second  root  is  essentially  negative. 

Hence,  in  the  third  form,  both  roots  are  negative.  See  problems 
12  and  13,  Art.  212. 

In  the  equation  x2 — 2px= — g,  or  fourth  form,  the  two  values 
of  x  are 

+P+V-q+p2 

And  -\-p— V— Q+P2- 

The  value  of  the  radical  part,  as  in  the  preceding  form,  is  less 
than  p.     Hence,  the  first  root  is  essentially  positive. 

The  second  root,  being  equal  to  p,  minus  a  quantity  less  than  p, 
is  essentially  positive. 

Hence,  in  the  fourth  form,  both  roots  are  positive.  See  prob- 
lems 14  and  15,  Art.  212. 

Review. — 215.  Show  that  every  quadratic  equation  has  two  roots, 
and  only  two.  216.  To  what  is  the  sum  of  the  roots  equal?  To  what 
is  the  product  equal? 

217.  Show  that  in  the  first  form  one  of  the  roots  is  positive,  and 
the  other  negative ;  and  that  the  negative  root  is  numerically  greater 
than  the  positive. 


208  RAY  S  ALGEBRA,  FIRST  BOOK. 

218.  In  the  third  and  fourth  forms,  the  radical  part, 
|/ — q-j-p2,  will  be  further  considered. 

If  q  is  greater  than  p2,  we  are  required  to  extract  the  square  root 
of  a  negative  quantity,  which  is  impossible.  See  Art.  195.  There- 
fore, 

In  the  third  and  fourth  forms,  when  q  is  greater  than  p2,  that  is, 
when  the  known  term  is  negative,  and  greater  than  the  square  of 
half  the  coefficient  of  the  first  power  of  X,  both  values  of  the  un- 
known quantity  are  impossible. 

To  explain  the  cause  of  this  impossibility,  we  must  first  solve  the 
following  problem: 

Into  what  two  parts  must  a  number  be  divided,  so  that 
the  product  of  the  parts  shall  be  the  greatest  possible  ? 

Let  2p  represent  any  number,  and  let  the  parts,  into  which  it  is 
supposed  to  be  divided,  be  p-\-z  and  p — z.  The  product  of  these 
parts  is  (p-\-z)(p—z)=p2—z2. 

Now,  this  product  increases  as  z  diminishes,  and  is  greatest  when 
Z2  is  least;  that  is,  when  z2  or  z  is  0.  But,  when  z  is  0,  the  parts 
are  p  and  p.     Hence, 

When  a  number  is  divided  into  two  equal  parts,  their 
product  is  greater  than  that  of  any  other  two  parts  into 
which  it  can  be  divided. 

Or,  as  the  same  principle  may  be  otherwise  expressed, 

The  product  of  any  two  unequal  numbers  is  less  than  the 
square  of  half  their  sum. 

As  an  illustration  of  this  principle,  take  the  number  10,  and 
divide  it  into  different  parts. 

10=9+1,  and  9x1=  9 
10=8+2,  and  8x2=16 
10=7+3,  and  7x3=21 
10=6+4,  and  6x4=24 
10=5+5,  and  5X&=25 


QUADRATIC  EQUATIONS.  209 

We  see  that  the  product  of  the  parts  becomes  greater  as  they  ap- 
proach to  equality,  and'  greatest  when  they  are  equal. 

Now,  in  Art.  216,  it  has  been  shown  that  2p  is  equal  to  the  sum 
of  the  two  values  of  x,  and  that  q  is  equal  to  their  product.  But, 
when  q  is  greater  than  p2,  we  have  the  product  of  two  numbers 
greater  than  the  square  of  half  their  sum,  which  is  impossible. 

If,  then,  any  problem  furnishes  an  equation  in  which  the  known 
term  is  negative,  and  greater  than  the  square  of  half  the  coefficient 
of  x,  we  may  infer  that  the  conditions  of  the  problem  are  incom- 
patible with  each  other.     The  following  is  an  example ; 

Let  it  be  required  to  divide  the  number  12  into  two 
such  parts,  that  their  product  shall  be  40. 

Let  X  and  12 — x  represent  the  parts. 
Then,  ar(12— «)=40. 

Or,  x2—12x=— 40. 

a;2— 12z+36=— 4. 

x— 6=  =tyHl" 

And  a=6±i/=4~ 

These  values  show  that  the  problem  is  impossible.  This  we  also 
know,  from  the  preceding  theorem,  since  12  can  not  be  divided  into 
any  two  parts,  whose  product  will  be  greater  than  36. 

Remarks. — 1.  When  the  coefficient  of  x2  is  negative,  as  in  the 
equation  — x2-\-mx=n,  the  pupil  may  not  perceive  that  it  is  em- 
braced in  the  four  general  forms.  This  apparent  difficulty  is  ob- 
viated, by  multiplying  both  sides  of  the  equation  by  — 1,  or  by 
changing  the  signs  of  all  the  terms. 

2.  Since  the  sign  of  the  square  root  of  x2,  or  of  (X-\-p)2,  is  =b,  it 
might  seem,  that  when  #2=m2,  we  should  have  dLZX=zhm,  that 
is,  -\-x=-±im,  and  — x—-±zm-}  such  is  really  the  case,  but  — #=-|-m 
is  the  same  as  -J-aP= — m,  and  —  x= — m  is  the  same  as  -\-x=-\-m. 

Hence,  -f  a;=±m,  embraces  all  the  values  of  X.  In  the  same 
manner,  it  is  necessary  to  take  only  the  plus  sign  of  the  square  root 
of  (z-j-p)2. 

Review. — 217.  Show  that  in  the  2d  form,  one  root  is  positive  and 
the  other  negative;  and  that  the  positive  root  is  greater  than  the 
negative.  That  in  the  3d,  both  are  negative.  That  in  the  4th,  both 
are  positive. 

1st  Bk.  18 


210  RAYS  ALGEBRA,  FIRST  BOOK. 


QUADRATIC    EQUATIONS    CONTAINING    TWO 
UNKNOWN    QUANTITIES. 

Note. — A  full  discussion  of  equations  of  this  class  does  not 
properly  belong  to  an  elementary  treatise.  The  following  examples 
embrace  such  only  as  are  capable  of  solution  by  simple  methods. 
See  Ray's  Algebra,  Second  Bcok. 


219.  In  solving  equations  of  this  kind,  the  first  step 
is  to  eliminate  one  of  the  unknown  quantities.  This  may 
be  performed  by  methods  already  given.  See  Arts.  158. 
159,  160. 

1.  Given  x — y=2  and  x2-\-y2—100,  to  find  x  and  y. 
By  the  first  equation,  x=y-\-2. 

Substituting  this  value  of  x,  in  the  second, 

(y+2)2+y2=100. 
From  which  we  readily  find,         y=6,  or  — 8. 
Hence,  x=y-\-2=8,  or  — 6. 

2.  Given  x-\-y=S,  and  cey=15,  to  find  x  and  y. 

From  the  first  equation,  x=8 — y. 

Substituting  this  value  of  x,  in  the  second, 

y(8-y)=l5. 
Or,  y2—8y=—15. 

From  which  y  is  found  to  be  5  or  3. 
Hence,  x=S,  or  5. 

There  is  a  general  method  of  solving  questions  of  this  form,  with- 
out completing  the  square,  which  should  be  well  understood.  To 
explain  it,  suppose  we  have  the  equations, 

xA-y^a. 
xy=b. 
Squaring  the  first,        x2-{-2xy-]-y2—a2. 
Multiplying  the  second  by  4,    4xy=4b. 

Subtracting,  x2 — 2xy-{-  y2=a2—  46. 


QUADRATIC  EQUATIONS.  211 


Extracting  the  square  root,  x — y=dzy/ a2 — 46. 

But  x-\-y=a. 

Adding,  2x=adzl/a2—  46. 

Or,  x=la±hi/a2—4b. 

Subtracting,  2?/=azp:y/a2— 46. 

Or,  y^\a^z\^/a2— 46. 

If  we  have  the  equations  X — y=a  and  xy=b,  we  may  find  the 
values  of  x  and  y,  in  like  manner,  by  squaring  each  member  of  the 
first  equation,  and  adding  to  each  side  4  times  the  second. 

Extracting  the  square  root,  we  obtain  the  value  of 


iC+?/=±v/a2+46. 
From  this,  and  X — y=ct, 


We  find  x=\a±i\^/  a2+\b. 

2/=|azFi1/a2-f46. 

3.  Given  x-\-y=a  and  x2-\-y2=b,  to  find  x  and  y. 

Squaring  the  first,  x2-\-2xy-\-y2=a2  (3) 

But  x2-\-y2—b  (2) 

Subtracting,  2xy=a2 — 6      (4) 

Take  (4)  from  (2),  x2—2xy-\-y2=2b—a2. 

Extracting  the  root,  x — 2/=±y/26 — a2. 

x-{-y=a. 


Adding  and  dividing,  x=^adz^2b — a2. 

Subtracting  and  dividing,  y=£aqp£|/26 — a2. 


4.  Given  x2-{-y2=a  and  xy=b,  to  find  x  and  y. 
Adding  twice  the  second  to  the  first, 

x2-\-2xy+y2=a-\-2b. 
Extracting  the  square  root,         x-\-y=dzy  a-\-2b. 

Subtracting  twice  the  second  from  the  first, 

x2 — 2xy-\-y2=a— 26. 
Extracting  the  square  root,         x — y=d=l/a — 26. 


Whence,  a— ri^a-f  26dr|l/a— 26. 

And  y=zt2%ya-\-2bzptya— 26. 


212 


RAY'S  ALGEBRA,  FIRST  BOOK. 


5.  Given  x*-{-yz==a  and  x+y=b,  to  find  x  and  y. 

Dividing  the  first  by  the  second, 

x2-xy+y2=^  (3) 

Squaring  the  second,       x2-\-2xy-\-y2=b2.         (4) 

53 a 

Subtracting  (3)  from  (4),  Sxy 


Or, 

Take  (5)  from  (3), 

Extracting  the  root, 

But 

Whence, 


6 
b3—a   ,« 

=-o7T-    (5) 


x2—2xy+y2z 


'   36 
4a— & 
~~3b~' 


And 


x+y=b. 

4a— 63  \ 


36 


*V( 


36 


=a  a 
)+ J6,  and  y=±l^  (  4-^  )-Ji. 


In  a  similar  manner,  if  we  have  Xz — y^=a  and  X — y=b}  we  find 
4a — 63  \    . .        .  .  .    /  /  4a — 63 


6.  a)2+,y2r=34| 

7.  a+^16} 

xy  =63  j 

8.  x  — y  =   5  I 

ay  =36  j 

9.  x  -\-y 

10.  x— y=    5) 

a;2-]-3/2=73  j 


:9) 

53  j 


11.  a»+y,=152| 

ic  -{-y  sac      8  j 


.    Ans.  x=zb5. 
.    .     .  y=±3. 

Ans.  £C=9,  or  7. 
y=7,  or  9. 

Ans.  x=9,  or  — 4. 
.  y=4,  or  — 9. 

.ns.  x=l,  or  2. 
.     .  y=2,  or  7. 

Ans.  rc=8,  or  ■ — 3. 
.  y=3,  or  — 8. 

Ans.  x=S,  or  3. 
.     .y=3,  or  5. 


EQUATIONS  OF  THE  SECOND  DEGREE. 


213 


12.  xs- ys=20S) Ans.  x 

x—y  =     4j y 

13.  *3+*/3=190r-fy)  |     . 
x — y=3  j 

14.  x  A-v  =11  ) Ans.  x 


+y=ll)      .     .     .  . 

.     .     .  . 

=12)  . 

=12  j  . 


15.  (*-3)(H-2)=12 

xy= 

16.  y— x=2 

Bxt/="L0x-j-y 


17.  3xs+2xy=24 
5x  — 3y 


11: 


18. 


1     1 


x     y 

V-1 


—  -J— J  3 

^H"     2 3g 


19.  a— y=2 

x2y2=21—4xy 


6,  or  —2. 
2,  or  —6. 


.    Ans.  ic=-5,  or  — 2. 
.     .     .  y=2,  or  —5. 


y=5. 

.    Ans.  #=6,  or  — 3. 
.     .     .  y=2,  or  —4. 

.    Ans.  x=2,  or  — J. 
.     .     .    ?/-_4,  or    If. 

.  Ans.  x=2,  or  — 1|. 


3      nr   I  9  9 

o,  or         g7-. 


Ans.  x=2,  or  3. 

.     .  y__3,  or  2. 


Ans.  se=3,  or  — 1. 
3. 


:1,  or 


In  solving  question  18,  let  — =v,  and  -— z;  the  question  then  be- 
x  y 

comes  similar  to  the  9th.     In  question  19,  find  the  value  of  xy  from 

the  second  equation,  as  if  it  were  a  single  unknown  quantity. 


TROBLEMS    PRODUCING    QUADRATIC    EQUATIONS 
CONTAINING    TWO    UNKNOWN    QUANTITIES. 

1.  The  sum  of  two  numbers  is  10,  and  the  sum  of  their 
squares  52  ;  what  are  the  numbers?  Ans.  4  and  6. 

2.  The  difference  of  two  numbers  is  3,  and  the  differ- 
ence of  their  squares  39;  required  the  numbers. 

Ans.  8  and  5. 


214  RAY'S  ALGEBRA,  FIRST  BOOK. 

3.  Divide  the  number  25  into  two  such  parts,  that  the 
sum  of  their  square  roots  shall  be  7.      Ans.  16  and  9. 

4.  The  product  of  a  number,  consisting  of  two  places, 
by  the  sum  of  its  digits,  is  160  ;  divided  by  4  times  the 
digit  in  unit's  place,  the  quotient  is  4  ;  what  is  the  num- 
ber? Ans.  32. 

5.  The  difference  of  two  numbers,  multiplied  by  the 
greater,  =16,  but  by  the  less,  =12;  required  the  num- 
bers. Ans.  8  and  6 

6.  Divide  10  into  two  such  parts,  that  their  product 
shall  exceed  their  difference  by  22.  Ans.  6  and  4. 

7.  The  sum  of  two  numbers  is  10,  and  the  sum  of  their 
cubes  is  370  ;  required  the  numbers.        Ans.  3  and  7. 

8.  The  difference  of  two  numbers  is  2,  and  the  differ- 
ence of  their  cubes  is  98  ;  required  the  numbers. 

Ans.  5  and  3. 

9.  If  a  number,  consisting  of  two  places,  is  divided  by 
the  product  of  its  digits,  the  quotient  will  be  2 ;  if  27  is 
added  to  it,  the  digits  will  be  inverted ;  what  is  the  num- 
ber? Ans.  36. 

10.  Find  three  such  quantities,  that  the  quotients  arising 
from  dividing  the  products  of  every  two  of  them,  by  the 
one  remaining,  are  a,  h,  and  c. 

Ans.  zh-y/ab,  dbj/ac,  and  zbj/6c. 

11.  Find  two  numbers,  the  sum  of  whose  squares  ex- 
ceeds twice  their  product,  by  4,  and  the  difference  of  whose 
squares  exceeds  half  their  product,  by  4.     Ans.  6  and  8. 

12.  The  fore  wheel  of  a  carriage  makes  6  revolutions 
more  than  the  hind  wheel,  in  going  120  yd. ;  but  if  the 
circumference  of  each  wheel  is  increased  1  yd.,  it  will 
make  only  4  revolutions  more  than  the  hind  wheel,  in  the 
same  distance ;  required  the  circumference  of  each  wheel. 

Ans.  4  and  5  yd. 


QUADRATIC  EQUATIONS.  215 

13.  A  and  B  depart  from  the  same  place,  and  travel  in 
the  same  direction  ;  A  starts  2  hr.  before  B,  and  after 
traveling  30  mi.,  B  overtakes  A  ;  had  each  traveled  half 
a  mi.  more  per  hr.,  B  would  have  traveled  42  mi.  before 
overtaking  A.     At  what  rate  did  they  travel  ? 

Ans.  A.  2£,  B  3  mi.  per  hr. 

14.  A  and  B  started  at  the  same  time,  from  two  differ- 
ent points,  toward  each  other ;  when  they  met  on  the  road, 
it  appeared  that  A  had  traveled  30  mi.  more  than  B.  It 
also  appeared  that  it  would  take  A  4  da.  to  travel  the  road 
that  B  had  come,  and  B  9  da.  to  travel  the  road  that  A 
had  come.     Find  the  distance  of  A  from  B  at  starting. 

Ans.  150  mi. 


GENERAL    REVIEW. 

Define  algebra.  Unit  of  measure.  Difference  between  an  ab- 
stract and  concrete  number.  Between  a  theorem  and  a  problem. 
Define  power  of  a  quantity.  Coefficient.  Exponent.  Root,  Recip- 
rocal of  a  quantity.  Subtraction.  How  does  algebraic  differ  from 
arithmetical  subtraction?     Illustrate. 

Difference  between  a  prime  and  a  composite  number.  When  are 
two  quantities  prime  to  each  other?  Define  the  G.C.D.  of  two  or 
more  numbers.  The  L.C.M.  A  fraction.  Terms  of  a  fraction. 
How  add  fractions?     Multiply?     Divide? 

Define  an  equation.  How  many  classes  of  quantities  in  an  equa- 
tion? Define  a  quadratic  equation.  A  literal  equation.  How  clear 
an  equation  of  fractions?  Rule  for  the  solution  of  simple  equations. 
Define  elimination.  Rule  for  elimination  by  substitution.  By  addi- 
tion and  subtraction.     By  comparison. 

Define  transposition.  How  are  signs  affected  by  transposition  ? 
How  many  methods  of  elimination  ?  Illustrate  each  by  an  example. 
Rule  for  elimination  in  three  or  more  unknown  quantities.  What 
is  meant  by  generalization?  Illustrate  by  an  example.  When  is 
the  answer  to  a  problem  termed  a  formula  ? 

What  is  a  negative  solution?  When  is  an  equation  independent? 
When  redundant?  Define  evolution.  What  is  the  square  root  of  a 
number?  Rule  for  extracting  square  root.  Why  can  not  a  binomial 
be  a  perfect  square?  Define  a  radical  of  the  second  degree.  A 
surd.     What  are  similar  radicals? 

Rule  for  addition  of  radicals.  Division.  State  difference  between 
a  pure  and  an  affected  quadratic  equation.  Rule  for  solution  of  an 
affected  quadratic.  Show  that  every  quadratic  equation  has  two 
roots,  and  only  two. 


216  RAY'S  ALGEBRA,  FIRST  BOOK. 


VIII.    PROGRESSIONS    AND 
PROPORTION. 

ARITHMETICAL    PROGRESSION. 

220.  A  Series  is  a  succession  of  quantities  or  numbers, 
connected  together  by  the  signs  +  or  — ,  in  which  suc- 
ceeding terms  may  be  derived  from  those  which  precede 
them,  by  a  rule  deducible  from  the  law  of  the  series. 

Thus,  1  +  3+5  +  7+9+,  etc., 

2+6+18+54+,    etc.,  are  series. 

In  the  former,  any  term  may  be  derived  from  that  which 
precedes  it,  by  adding  2  ;  and,  in  the  latter,  any  term  may 
be  found  by  multiplying  the  preceding  term  by  3. 

221.  An  Arithmetical  Progression  is  a  series  of 
quantities  which  increase  or  decrease,  by  a  common  differ- 
ence. 

Thus,  the  numbers  1,  3,  5,  7,  9,  etc.,  form  an  increas- 
ing arithmetical  progression,  in  which  the  common  differ- 
ence is  2. 

The  numbers  30,  27,  24,  21,  18,  15,  etc.,  form  a 
decreasing  arithmetical  progression,  in  which  the  common 
difference  is  3. 

Remark. — An  arithmetical  progression  is  termed,  by  some  writ- 
ers, an  equidifferent  series,  or  a  progression  by  difference. 

Again,  a,  a-\-d,  a-\-2d,  a-\-Sd,  a+4d,  a+5c7,  etc.,  is 
an  increasing  arithmetical  progression,  whose  first  term 
is  a,  and  common  difference  d. 

If  d  be  negative,  it  becomes  a,  a — d,  a — 2d,  a — 3d, 
a — 4cZ,  a — 5d,  etc.,  which  is  a  decreasing  arithmetical  pro- 


ARITHMETICAL  PROGRESSION.  217 

2!22«  If  we  take  an  arithmetical  series,  of  which  the 
first  terra  is  a,  and  common  difference  d,  we  have 

1st  term  = a 

2d  term  =lst  term  -\-d—a-\-  d 
3d  term  =2d  term  -{-d=a-\-2d 
4th  term  =3d    term  -\-d=a-\- 3d;  and  so  on. 

Hence,  the  coefficient  of  d  in  any  term  is  less  by  unity, 
than  the  number  of  that  term  in  the  series ;  therefore,  the 
nth  term  =a-\-(n — l)d. 

If  we  designate  the  ?ith  term  by  ?,  we  have  l—a-\-(ii — V)d. 
For   a   decreasing  series   we  also  have   l=a — (n — l)a. 
Hence, 


Rule. — 1.  For  an  Increasing  Series. — Multiply  the 
common  difference  by  the  number  of  terms  less  one,  and  add 
the  product  to  the  first  term. 

2.  For  a  Decreasing  Series. — Multiply  the  common 
difference  by  the  number  of  terms  less  one,  and  subtract  the 
product  from  the  first  term. 

1.  The  first  term  of  an  increasing  arithmetical  series  is 
3,  and  common  difference  5  ;  required  the  8th  term. 

Here  I,  or  8th  term,  =3+ (8— 1)5=3+35=38,  Ans. 

2.  The  first  term  of  a  decreasing  arithmetical  series 
is  50,  and  common  difference  3  ;  required  the  10th  term. 

Here  I,  or  10th  term,  =50— (10— 1)3=50— 27=23,  Ans. 

Review. — 220.  What  is  a  series?  221.  What  is  an  arithmetical 
progression?  Give  an  example  of  an  increasing  series.  Of  a  de- 
creasing series. 

222.  Rule  for  finding  the  last  term  of  an  increasing  arithmetical 
series.     Of  a  decreasing  series.     Prove  these  rules. 

IstBk.  10* 


218  RAYS  ALGEBRA,  FIRST  BOOK. 

In  the  following  examples,  a  denotes  the  first  term,  and  d  the 
common  difference  of  an  arithmetical  series;  d  being  plus  when 
the  series  is  increasing,  and  minus  when  it  is  decreasing. 

3.  a=3,  and  d=&  ;  required  the  6th  term.         Ans.  28. 

4.  a=7j  and  d=±  ;  required  the  16th  term.      Ans.  lOf. 

5.  a=2|,  and  cZ=i ;  required  the  100th  term.    Ans.  35J. 

6.  a=0,  and  d=%  ',  required  the  11th  term.         Ans.  5. 

7.  a=30,  and  d= — 2  ;  required  the  8th  term.     Ans.  16. 

8.  a= — 10,  and  d= — 2  ;  required  the  6th  term. 

Ans.  —20. 

9.  If  a  body  falls  during  20  sec.,  descending  16^  ft. 
the  first  sec,  48]  ft.  the  next,  and  so  on,  how  far  will  it 
fall  the  twentieth  sec?  Ans.  627]  ft. 

223.  Given,  the  first  term  a,  the  common  difference  d, 
and  the  number  of  terms  n,  to  find  s,  the  sum  of  the  series. 

If  we  take  an  arithmetical  series,  of  which  the  first  term  is  3,  com- 
mon difference  2,  and  number  of  tei-ms  5,  it  may  be  written  in  the 
following  forms : 

3,  5,  7,  9,  11, 

11,  9,  7,  5,  3. 

It  is  obvious  that  the  sum  of  all  the  terms  in  either  of  these  lines 
will  represent  the  sum  of  the  series ;  that  is, 

s=  3+  5+  7+  9+11 
And  s=ll-f  9-j-  7-f-  5+  3 

Adding,  2s=14+14+14+14+14 

=14X5,  the  number  of  terms,  =70. 
Whence,  s=%  of  70=35. 

Now,  let  ?=  the  last  term,  and  n=  the  number  of  terms.  Writ- 
ing the  series  as  before, 

s=a+(a+d)+ (a+2d)+(a-\-3d)+  .    .    .  -f  I 
And  s=l+(l—  d)+(l—  2d)+\l—  3d)+  .     .    .  -fa 

Adding,    2s=(Z+a)-f-(Z+a)-f  (£-j- a)-f  (£+ a)      .    .    .  -\-(l+a) 
Hence,      2s=(l+a)n,  and 

*=(?-f  a)^=  (  -±^  y.     Hence, 


ARITHMETICAL  PROGRESSION.  219 


Rule. — Multiply  half  the  sum  of  the  two  extremes  by  the 
number  of  terms. 

From  the  preceding,  it  appears  that  the  sum  of  the  ex- 
tremes is  equal  to  the  sum  of  any  other  two  terms  equally 
distant  from  the  extremes. 

Since  l=a-\-(n — l)d,  if  we  substitute  this  in  the  place  of 

ill 
I  in  the   formula  s  =  (I  -j-  a)  -,  it   becomes   s  =  (2a  -f- 

(n  —  1)  d)  -'     Hence, 
2S 


TO    FIND    THE    SUM    OF    AN    ARITHMETICAL    SERIES, 

Rule. —  To  twice  the  first  term,  add  the  product  of  the 
number  of  terms  less  one,  by  the  common  difference,  and  mul- 
tiply the  sum  by  half  the  number  of  terms. 

1.  Find  the  sum  of  an  arithmetical  series,  of  which  the 
first  term  is  3,  last  term  17,  and  number  of  terms  8. 


/H±>\e_ 


-m 


8=80,  Ans. 


2.  Find  the  sum  of  an  arithmetical   series,   whose  first 
term  is  1,  last  term  12,  and  number  of  terms  12. 

Ans.  78. 

3.  Find  the  sum  of  an  arithmetical  series,  whose  first 
term  is  0,  common  difference  1,  and  number  of  terms  20. 

Ans.  190. 

4.  Find  the  sum  of  an   arithmetical  series,  whose  first 
term  is  3,  common  difference  2,  and  number  of  terms  21. 

Ans.  483. 

Review.  —223.  What  is  the  rule  for  finding  the  sum  of  an  arith- 
metical series?     Prove  the  rule. 


220  RAY'S  ALGEBRA,  FIRST  BOOK. 

5.  Find  the  sum  of  an  arithmetical  series,  whose  first 
term  is  10,  common  difference  — 3,  and  number  of  terms  10. 

Ans.  —35. 

224.  The  equations,  l=a-\-(n — V)d 

furnish  the  means  of  solving  this  general  problem : 

Knowing  any  three  of  the  five  quantities  a,  d,  n,  I,  s,  which 
enter  into  an  arithmetical  series,  to  determine  the  other  two. 

This  question  furnishes  ten  cases,  for  the  solution  of 
which  we  have  always  two  equations,  with  only  two  un- 
known quantities. 

1.  Let  it  be  required  to  find  a  in  terms  of  I,  n,  d. 

From  the  first  formula,  by  transposing,  we  have 
a=l — (n — l)cZ;  that  is, 

The  first  term  of  an  increasing  arithmetical  series  is  equal 
to  the  last  term  diminished,  by  the  product  of  the  common 
difference  into  the  number  of  terms  less  one. 

From  the  same  formula,  we  find  d= T ;  that  is, 

n — 1 

In  any  arithmetical  series,  the  common  difference  is  equal 
to  the  difference  of  the  extremes,  divided  by  the  number  of 
terms  less  one. 

223.  By  means  of  the  preceding  rules,  we  are  enabled 
to  solve  such  problems  as  the  following : 

Review. — 224.  What  are  the  fundamental  equations  of  arithmeti- 
cal progression,  and  to  what  general  problem  do  they  give  rise  ? 

224.  To  what  is  the  first  term  of  an  increasing  arithmetical  series 
equal  ?  To  what  is  the  common  difference  of  an  arithmetical  series 
equal  ? 


ARITHMETICAL  PROGRESSION.  221 

Let  it  be  required  to  insert  five  arithmetical  means  be- 
tween 3  and  15. 

Here,  the  two  given  terms  with  the  five  to  be  inserted  make  seven. 
Hence,  n=z7,  (X=3,  £=15,  from  which  we  find  d=2.  Adding  the 
common  difference  to  3  and  the  succeeding  terms,  we  have  for  the 
series  3,  5,  7,  9,  11,  13,  15. 

If  we  insert  the  same  number  of  means  between  the  consecutive 
terms  of  a  series,  the  result  will  form  a  new  progression.     Thus, 

If  we  insert  3  terms  between  the  terms  in  1,  9,  17,  etc.,  the  new 
series  will  be  1,  3,  5,  7,  9,  11,  13,  15,  17,  etc. 

1.  Find  tlie  sum  of  the  natural  series  of  numbers  1,  2, 
3,  4,    ...    .    carried  to  1000  terms.         Ans.  500500. 

2.  Required  the  last  term,  and  the  sum  of  the  series, 
1,  3,  5,  7,  ...   .  to  101  terms.     Ans.  201  and  10201. 

3.  How  many  times  does  a  common  clock  strike  in  a 
week?  Ans.  1092. 

4.  Find  the  nth.  term,  and  the  sum  of  n  terms  of  the 
natural  series  of  numbers  1,  2,  3,  4,    .    .    .    . 

Ans.  n)  and  ^n(n-\-Y). 

5.  The  first  and  last  terms  of  an  arithmetical  series 
are  2  and  29,  and  the  common  difference  is  3  ;  required 
the  number  of  terms  and  the  sum  of  the  series. 

Ans.  10  and  155. 

6.  The  first  and  last  terms  of  a  decreasing  arithmetical 
series  are  10  and  6,  and  the  number  of  terms  9  ;  required 
the  common  difference,  and  the  sum  of  the  series. 

Ans.  i  and  72. 

7.  The  first  term  of  a  decreasing  arithmetical  series 
is  10,  the  number  of  terms  10,  and  the  sum  of  the  series 
85  ;  required  the  last  term  and  the  common  difference. 

Ans.  7  and  J. 

8.  Required  the  series  obtained  from  inserting  four 
arithmetical  means  between  each  of  the  two  terms  of  the 
series  1,  16,  31,  etc.  Ans.  1,  4,  7,  10,  13,  16,  etc. 


222  RAY'S  ALGEBRA,  FIRST  BOOK. 

9.  The  sum  of  an  arithmetical  progression  is  ^2,  the  first 
term  is  24,  and  the  common  difference  is  — 4;  required  the 
number  of  terms.  Ans.  9,  or  4. 

This  question  presents  the  equation  n2 — 13n= — 36,  which  has  two 
roots,  9  and  4.  These  give  rise  to  the  two  following  series,  in  each 
of  which  the  sum  is  72. 

First  series,         24,  20,  16,  12,  8,  4,  0,  ~-4,  —8. 
Second  series,     24,  20,  16,  12. 

10.  A  man  bought  a  farm,  paying  for  the  first  A.  $1,  for 
the  second  $2,  for  the  third  $3,  and  so  on ;  when  he  came 
to  settle,  he  had  to  pay  $12880;  how  many  A.  did  the 
farm  contain,  and  what  was  the  average  price  per  A.  ? 

Ans.  160  A.,  at  $801  per  A. 

11.  If  A  start  from  a  certain  place,  traveling  a  mi.  the 
first  da.,  2a  the  second,  3a  the  third,  and  so  on ;  and  at  the 
end  of  4  da.,  B  start  after  him  from  the  same  place,  travel- 
ing uniformly  9a  mi.  a  da. ;  when  will  B  overtake  A  ? 

Let  x .=  the  number  of  da.  required ;  then,  the  distance  traveled 
by  A  in  X  da.  =a-f  2a-f3a,  etc.,  to  x  terms,  =\ax(x-\-l)\  and  the 
distance  traveled  by  B  in  (x — 4)  da,  =  9a(x — 4). 

Whence,  $ax(x-\-l)= 9a(x— 4).     From  which  x=8,  or  9. 

Hence,  B  overtakes  A  at  the  end  of  8  da.;  and  since,  on  the  ninth 
da.,  A  travels  9a  mi.,  which  is  B's  uniform  rate,  they  will  be  to- 
gether at  the  end  of  the  ninth  da. 

12.  A  sets  out  3  hr.  and  20  min.  before  B,  and  travels 
at  the  rate  of  6  mi.  an  hr. ;  in  how  many  hr.  will  B  over- 
take A,  if  he  travel  5  mi.  the  first  hr.,  6  the  second,  7  the 
third,  and  so  on  ?  Ans.  8  hr. 

13.  A  and  B  set  out  from  the  same  place,  at  the  same 
time.  A  travels  at  the  constant  rate  of  3  mi.  an  hr., 
but  B's  rate  of  traveling  is  4  mi.  the  first  hr.,  3i  the  sec- 
ond, 3  the  third,  and  so  on ;  in  how  many  hr.  will  A  over- 
take B?  Ans.  5  hr. 

Review. — 225.  How  do  you  insert  any  number  of  arithmetical 
means  between  two  given  numbers? 


GEOMETRICAL  PROGRESSION.  223 


GEOMETRICAL    PROGRESSION. 

226.  A  Geometrical  Progression  is  a  series  of  terms, 
each  of  which  is  derived  from  the  preceding,  by  multiply- 
ing it  by  a  constant  quantity,  termed  the  ratio. 

Thus,  1,  2,  4,  8,  16,  etc.,  is  an  increasing  geometrical 
series,  whose  common  ratio  is  2. 

Also,  54,  18,  6,  2,  etc.,  is  a  decreasing  geometrical  series, 
whose  common  ratio  is  £. 

Generally,  a,  ar,  ar2,  ar3,  etc.,  is  a  geometrical  progres- 
sion, whose  common  ratio  is  r,  and  which  is  an  increasing 
or  decreasing  series,  according  as  r  is  greater  or  less  than  1. 

It  is  obvious  that  the  common  ratio  will  be  ascertained 
by  dividing  any  term  of  the  series  by  that  which  pre- 
cedes it. 

227.  To  find  the  last  term  of  the  series. 

Let  a  denote  the  1st  term,  r  the  ratio,  I  the  Tith  term, 
and  s  the  sum  of  n  terms ;  then,  the  respective  terms  of 
the  series  will  be 

1,    2,      3,     4,      5    .     .     .     .     Ti—3,    n— 2,    n— 1,  n 

a,  ar,  ar2,  ar3,  ar4    ,    .     .    .    ar"-4,  ar"-3,  arn-\  arn-\ 

That  is,  the  exponent  of  r  in  the  second  term  is  1,  in  the 
third  2,  in  the  fourth  3,  and  so  on;  the  nth.  term  of  the 
series  will  be,  l=arn~1.     Hence, 


TO    FIND    ANY    TERM    OF    A    GEOMETRICAL    SERIES, 

Rule. — Multiply  the  first  term   by  the  ratio   raised  to  a 
power,  whose  exponent  is  one  less  than  the  number  of  terms. 

1.  Find  the  5th  term  of  the  geometric  progression,  whose 
first  term  is  4,  and  common  ratio  3. 

^4X34=4X81=324,  the  fifth  term. 


224  RAY'S  ALGEBRA,  FIRST  BOOK. 

2.  Find  the  6th  term  of  the  progression  2,  8,  32,  etc. 

Ans.  2048. 

3.  Given  the  1st  term  1,  and  ratio  2,  to  find  the  7th 
term.  Ans.  64. 

4.  Given  the  1st  term  4,  and  ratio  3,  to  find  the  10th 
term.  Ans.  78732. 

5.  Find  the  9th  term  of  the  series,  2,  10,  50,  etc. 

Ans.  781250. 
G.  Given  the  first  term  8,  and  ratio  ^,  to  find  the  15th 
term.  Ans.  ^W 

7.  A  man  purchased  9  horses,  agreeing  to  pay  for  the 
whole  what  the  last  would  cost,  at  $2  for  the  first,  $6  for 
the  second,  etc.  ;  what  was  the  average  price  of  each  ? 

Ans.  $1458. 

228.  To  find  the  sum  of  all  the  terms  of  the  series. 

Let  a,  ar,  ar2,  «r3,  etc.,  be  any  geometrical  series,  and  s  its  sum; 
then,  s=za-\-ar-\-ar2^-ar3 -f  arn-2-\-arn~l 

Multiplying  this  equation  by  r,  we  have 

rs=ar-f-ar2-f  ar3-|-ar4     ....     -far"-1 -far71. 

The  terms  of  the  two  series  are  identical,  except  the  first  term  of 
the  first  series,  and  the  last  term  of  the  second  series.  Subtracting 
the  first  equation  from  the  second,  we  have 

rs — s=arn — a 
Or,  (r— l)s=a(rn— 1) 

a(rn— 1) 
Hence,  8==    r—l 

Since  Zr^ar""1,  we  have    rl=arn 

arn—a    rl — a     TT 

Therefore,  s= =— = r.     Hence, 

'  r — 1        r — 1 

Review. — 226.  What  is  a  geometrical  progression  ?  Give  ex- 
ample of  an  increasing  geometrical  series.  Of  a  decreasing.  How 
may  the  common  ratio  in  any  geometrical  series  be  found? 

227.  How  is  any  term  of  a  geometrical  series  found?  Explain 
the  principle  of  this  rule. 


GEOMETRICAL  PROGRESSION.  225 


TO    FIND    THE    SUM    OF    A    GEOMETRICAL    SERIES, 

Rule. — Multiply  the  last  term  by  the  ratio,  from  the  prod- 
uct subtract  the  first  term,  and  divide  the  remainder  by  the 
ratio  less  one. 

1.  Find  the  sum  of  10  terms  of  the  progression  2,  6, 
18,  54,  etc. 

The  last  term  =2y39  =2  X  19683=39366. 

Ir—a    118098—2     _nn.0    , 

s— ^-= — = — = — =59048,  Ans. 

r  —  1  3—1 

2.  Find  the  sum  of  7  terms  of  the  progression  1,  2,  4, 
etc.  Ans.  127. 

3.  Find  the  sum  of  10  terms  of  the  progression  4,  12, 
3G,  etc.  Ans.  118096. 

4.  Find  the  sum  of  8  terms  of  the  series,  whose  first 
term  is  6{,  and  ratio  j.  Ans.  307|^. 

5.  Find  the  sum  of  3-f  4i+6|-f-,  etc.,  to  5  terms. 

Ans.  39TV 

If  the  ratio  T  is  less  than  1,  the  progression  is  decreasing,  and 
the  last  term  Ir  is  less  than  a.  To  render  both  terms  of  the  fraction 
positive,   change   the   signs  of  the  terms,   Art.  132,   and  we  have 

S=?  - ,   for  the  sum  of  the  series  when   the  progression  is  de- 
creasing. 

6.  Find  the  sum  of  15  terms  of  the  series  8,  4,  2,  1, 
etc.  Ans.  15f°|;|. 

7.  Find  the  sum  of  6  terms  of  the  series  6,  4^,  8 J, 
etc.  Ans.  19f||. 

Review. — 228.  Rule  for  the  sum  of  the  terms  of  a  geometrical 
series.  Prove  this  rule.  When  the  series  is  decreasing,  how  may 
the  formula  be  written  so  that  both  terms  of  the  fraction  may  be 
positive? 


226  RAY'S  ALGEBRA,  FIRST  BOOK. 

229.  In  a  decreasing  geometrical  series,  when  the 
number  of  terms  is  infinite,  the  last  term  becomes  in- 
finitely small,  that  is,  0.     Therefore,  Wr=0,  and  the  formula 

s=^ becomes  s=_, .     Hence, 

1 — r  1 — r 


TO    FIND    THE    SUM    OF   AN   INFINITE   DECREASING   SERIES, 
Rule. — Divide  the  first  term  by  one  minus  the  ratio. 

1.  Find  the  sum  of  the  infinite  series  1  +  J+^-f,  etc. 
Here,  a=l,  r=£,  and  s=^~ -=— ^=|=|,  Ans. 

2.  Find  the  sum  of  the  infinite  series  1-j-i-j-J-j-^-j-,  etc. 

Ans.  2. 

3.  Find  the  sum  of  the  infinite  series  9-f-6-{-4-{-,  etc. 

Ans.  27. 

4.  Find  the  sum  of  the  infinite  series  1 — ^-j-g — 3*7 +> 
etc.  Ans.  |. 

5.  Find  the  sum  of  the  infinite  series  l-\ — =-j — H — *4-. 

1  x2  '  #*  '  x*  ' 

etc.  x2 

Ans 


2— r 

6.  Find  the  sum  of  the  infinite  geometrical  progression 

7   ,  b2       b3     b*                 .    '  ■    '.  ■       .          .     .         b 
a — 6-| 2~r~^ — 3  etc-?  in  which  the  ratio  is . 


a       a'     or 


Ans. 


a+b' 

7.  A  body  moves  10  ft.  the  first  sec,  5  the  next,  2±  the 
next,  and  so  on  forever ;  how  many  ft.  would  it  move 
over?  Ans.  20. 

Review. — 229.  Rule  for  the  sum  of  a  decreasing  geometrical 
series,  when  the  number  of  terms  is  infinite.     Prove  this  rule. 


GEOMETRICAL  PROGRESSION.  227 

230.    The   equations,  l=arn~1,  and   s= ~    furnish 

r — 1 

this  general  problem : 

Knowing  three  of  (he  Jive  quantities  a,  r,  n,  I,  and  s,  of  a 
geometrical  progression,  to  find  the  other  two. 

This  problem  embraces  ten  different  questions,  as  in 
arithmetical  progression.  Some  of  these,  however,  involve 
the  extraction  of  high  roots,  the  application  of  logarithms, 
and  the  solution  of  higher  equations  than  those  treated  of 
in  the  preceding  pages. 

The  following  is  one  of  the  most  simple  and  useful  of 
these  cases : 

Having  given  the  first  and  last  terms,  and  the  number  of 
terms  of  a  geometrical  progression,  to  find  the  ratio. 

Here,  l=arn~\  or  rn~x=-.  Hence,  r=n~l^j  (~\. 

1.  The  first  and  last  terms  of  a  geometrical  series 
are  3  and  48,  the  number  of  terms  5  ;  required  the  inter- 
mediate terms. 

Here,  1=48,  a =3,  n— 1 =5—1=4. 

Hence,  r=V^==V  16=2. 

2.  In  a  geometrical  series  of  three  terms,  the  first  and 
last  terms  are  4  and  16;  required  the  middle  term. 

Ans.  8. 
In  a  geometrical  progression  of  three  terms,  the  middle  term  is 
called  a  mean  proportional  between  the  other  two. 

3.  Find  a  mean  proportional  between  8  and  32. 

Ans.  16. 

4.  The  first  and  last  terms  of  a  geometrical  series  are 
2  and  162,  and  the  number  of  terms  5  ;  required  the  ratio. 

Ans.  3. 


228  KAY'S  ALGEBRA,  FIRST  BOOK. 


RATIO    AND    PROPORTION. 

231.  Two  quantities  of  the  same  kind  may  be  compared 
in  two  ways  : 

1st.  By  finding  how  much  the  one  exceeds  the  other. 

2d.    By  finding  how  many  times  the  one  contains  the  other. 

If  we  compare  the  numbers  2  and  6,  by  the  first  method, 
we  say  that  2  is  4  less  than  6,  or  that  6  is  4  greater  than  2. 

If  we  compare  2  and  6  by  the  second  method,  we  say 
that  6  is  equal  to  three  times  2,  or  that  2  is  one  third  of  6. 

The  second  method  of  comparison  gives  rise  to  propor- 
tion. 

232.  Ratio  is  the  quotient  which  arises  from  dividing 
one  quantity  by  another  of  the  same  land. 

Thus,  the  ratio  of  2  to  6  is  3  ;  the  ratio  of  a  to  ma  is  m. 

Remarks. — 1st.  In  comparing  two  numbers  or  quantities  by  their 
quotient,  the  number  expressing  the  ratio  will  depend  on  which  is 
made  the  standard  of  comparison.  Some  writers  make  the  first  of 
the  two  numbers  the  standard  of  comparison,  and  say  the  ratio 
of  2  to  6  is  3;  others  make  the  last  the  standard,  and  say  the  ratio 
of  2  to  6  is  ^.     The  former  method  is  adopted  in  this  work. 

2d.  In  order  that  two  quantities  may  have  a  ratio  to  each  other, 
they  must  be  of  the  same  kind.  Thus,  2  yd.  has  a  ratio  to  6  yd.,  be- 
cause the  latter  is  three  times  the  former;  but  2  yd.  has  no  ratio 
to  $6,  since  the  one  can  not  be  either  greater,  less,  or  any  number 
of  times  the  other. 

233.  When  two  numbers,  as  2  and  6,  are  compared, 
the  first  is  called  the  antecedent,  the  second  the  consequent. 

Review. — 231.  In  how  many  ways  may  two  quantities  of  the  same 
kind  be  compared?  Compare  2  and  6  by  the  first  method.  By  the 
second. 

232.  What  is  ratio?  Give  an  illustration.  233.  When  two  num- 
bers are  compared,  what  is  the  first  called?    The  second?    Example. 


RATIO  AND  PROPORTION.  229 

An  antecedent  and  consequent,  when  spoken  of  as  one, 
are  called  a  couplet;  when  spoken  of  as  two,  the  terms  of 
the  ratio. 

Thus,  when  the  ratio  of  2  to  6  is  spoken  of,  2  and  6  to- 
gether, form  a  couplet,  of  which  2  is  the  first  term,  and 
6  the  second. 

234.  Ratio  is  expressed  in  two  ways  : 

1st.  In  the  form  of  a  fraction,  of  which  the  antecedent  is 
the  denominator,  and  the  consequent  the  numerator. 

Thus,  the  ratio  of  2  to  6,  is  expressed  by  | ;  the  ratio 
of  3  to  12,  by  -y-,  etc. 

2d.  By  placing  two  points  (:)  between  the  terms. 
Thus,  the   ratio   of  2   to  6,  is  written    2:6;  the  ratio 
of  3  to  8,  3  :  8,  etc. 

235.  The  ratio  of  two  quantities  may  be  either  a  whole 
number,  a  common  fraction,  or  an  interminate  decimal. 

Thus,  the  ratio  of  2  to  6  is  |,  or  3. 

The  ratio  of  10  to  4  is  JL  or  %, 

10  5 

The  ratio  of  2  to  j/IFis  ^-,  or    '  '        ,  or  1.118+ . 

From  the  last  illustration,  it  is  obvious  that  the  ratio  of  two 
quantities  can  not  always  be  expressed  exactly,  except  by  sym- 
bols; but,  by  employing  decimals,  we  may  find  the  approximate  ratio 
to  any  required  degree  of  exactness. 

236.  Since  the  ratio  of  two  numbers  is  expressed  by  a 
fraction,  it  follows  that  whatever  is  true  of  a  fraction,  is 
true  of  the  terms  of  a  ratio.     Hence, 

Review. — 234.  When  are  the  antecedent  and  consequent  of  a 
ratio  called  a  couplet?  When  called  terms?  By  what  two  methods 
is  ratio  expressed  ?  Example.  235.  What  forms  may  the  ratio  of 
two  quantities  have? 


230  RAY'S  ALGEBRA,  FIRST  BOOK. 

1st.  To  multiply  the  consequent,  or  divide  the  antecedent 
by  any  number,  multiplies  the  ratio  by  that  number.     Arts. 

122,  125. 

Thus,  the  ratio  of  4  to  12,  is  3. 

The  ratio  of  4  to  12X&,  is  3x5. 

The  ratio  of  4-^2  to  12,  is  6,  which  is  equal  to  3x2. 

2d.  To  divide  the  consequent,  or  multiply  the  antecedent  by 
any  number,  divides  the  ratio  by  that  number.  Arts.  123, 
124. 

Thus,  the  ratio  of  3  to  24,  is  8. 

The  ratio  of  3  to  24-=-2,  is  4,  which  is  equal  to  8-=-2. 

The  ratio  of  3x2  to  24,  is  4,  which  is  equal  to  8-f-2. 

3d.  To  multiply  or  divide  both  the  antecedent  and  conse- 
quent by  any  number,  does  not  alter  the  ratio.     Arts.  126, 

127. 

Thus,  the  ratio  of  6  to  18,  is  3. 
The  ratio  of  6x2  to  18x2,  is  3. 
The  ratio  of  6--2  to  18--2,  is  3. 


2S7.  When  the  two  numbers  are  equal,  the  ratio  is 
called  a  ratio  of  equality ;  when  the  second  is  greater  than 
the  first,  a  ratio  of  greater  inequality;  when  less,  a  ratio  of 
less  inequality. 

Thus,  the  ratio  of  4  to  4,  is  a  ratio  of  equality. 
The  ratio  of  4  to  8,  is  a  ratio  of  greater  inequality. 
The  ratio  of  4  to  2,  is  a  ratio  of  less  inequality. 

We  see,  from  this,  that  a  ratio  of  equality  may  be  ex- 
pressed by  1  ;  a  ratio  of  greater  inequality,  by  a  number 

Review  —236  How  is  a  ratio  affected  by  multiplying  the  conse- 
quent, or  dividing  the  antecedent?  Why?  By  dividing  the  conse- 
quent, or  multiplying  the  antecedent?  Why?  By  multiplying;  or 
dividing  both  antecedent  and  consequent  by  the  same  number  ? 
Why? 

237  What  is  a  ratio  of  equality?  Of  greater  inequality?  Of 
less  inequality?     Examples. 


RATIO  AND  PROPORTION.  231 

greater  than  1  ;  and  a  ratio  of  less  inequality,  by  a  number 
less  than  1. 

238.  A  Compound  Ratio  is  the  product  of  two  or 
more  ratios. 

Thus,  the  ra-iio  L0,  compounded  with  the  ratio  &,  is  Wy(£=&Q—4:. 
In  this  case,  3  multiplied  by  5,  is  said  to  have  to  10  multiplied 
by  6,  the  ratio  compounded  of  the  ratios  of  3  to  10  and  5  to  6. 

239.  Ratios  may  be  compared  by  reducing  the  fractions 
which  represent  them  to  a  common  denominator. 

Thus,  the  ratio  of  2  to  5  is  less  than  the  ratio  of  3  to  8,  for  |  or 
*g5  is  less  than  |  or  L6. 

PROPORTION. 

240.  Proportion  is  an  equality  of  ratios. 

Thus,  if  a,  6,  c,  d  are   four  quantities,  such   that    -   is 

d  a 

equal  to  -,  then  a,  b,  c,  d  form  a  proportion,  and  we  say 

that  a  is  to  6,  as  c  is  to  d)  or,  that  a  has  the  same  ratio 
to  6,  that  c  has  to  d. 

Proportion  is  written  in  two  ways,  by  using, 

1st.  The  colon  and  double  colon  ;  thus,  a  :  b  :  :  c  :  d. 

2d.    The  sign  of  equality  ;  thus,  a  :  b  =  c  :  d. 

The  first  is  read,  a  is  to  b  as  c  is  to  d;  the  second  is 
read,  the  ratio  of  a  to  6  equals  the  ratio  of  c  to  d. 

From  the  preceding  definition,  it  follows,  that  when  four 
quantities  are  in  proportion,  the  second  divided  by  the  first 
gives  the  same  quotient  as  the  fourth  divided  by  the  third. 

Review. — 238.  When  are  two  or  more  ratios  said  to  be  com- 
pounded ?     Examples. 

239.  How  may  ratios  be  compared  to  each  other?  240.  What  is 
proportion?  Example.  How  are  four  quantities  in  proportion 
written?     How  read?     Examples. 


232  RAY'S  ALGEBRA,  FIRST  BOOK. 

This  is  the  test  of  the  proportionality  of  four  quantities. 
Thus,  if  a,  b,  c,  d  are  the  four  terms  of  a  true  proportion, 

so  that  a  :  b  :  :  c  :  d,  we  must  have  -=-. 

a      c 

If  these  fractions  are  equal  to  each  other,  the  proportion 
is  true;  if  they  are  not  equal,  it  is  false. 

Let  it  be  required  to  find  whether  3  :  8  :  :  2  :  5. 
Since  §==§  is  not  a  true  equation,  the  proportion  is  false. 

Remark. — The  words  ratio  and  proportion  are  often  misapplied. 
Thus,  two  quantities  are  said  to  be  in  the  proportion  of  3  to  4,  in- 
stead of,  in  the  ratio  of  3  to  4. 

A  ratio  subsists  between  two  quantities,  a  proportion  only  between 
four.     It  requires  two  equal  ratios  to  form  a  proportion. 

241.  In  the  proportion  a  :  b  :  :  c  :  d,  each  of  the  quan- 
tities a,  6,  c,  d  is  called  a  term.  The  first  and  last  terms 
are  called  the  extremes;  the  second  and  third,  the  means. 

242.  Of  four  proportional  quantities,  the  first  and  third 
are  called  antecedents;  and  the  second  and  fourth,  conse- 
quents, Art.  233.  The  last  is  said  to  be  a  fourth  propor- 
tional to  the  other  three,  taken  in  their  order. 

243.  Three  quantities  are  in  proportion,  when  the  first 
has  the  same  ratio  to  the  second  that  the  second  has  to  the 
third.  In  this  case,  the  middle  term  is  called  a  mean  pro- 
portional between  the  other  two.  Thus,  if  we  have  the 
proportion 

a  :  b  :  :  b  :  c 

then  b  is  called  a  mean  proportional  between  a  and  c,  and 
c  is  called  a  third  proportional  to  a  and  b. 

Review. — 240.  Give  examples  of  a  true  and  false  proportion. 
What  is  a  test  of  the  proportionality  of  four  quantities  ?  241.  What 
are  the  first  and  last  terms  of  a  proportion  called?  The  second 
and  third? 

242.  What  are  the  first  and  third  terms  of  a  proportion  called  ? 
The  second  and  fourth  ?  243.  When  are  three  quantities  in  propor- 
tion?    Example.     What  is  the  second  term  called ?     The  third? 


RATIO  AND  PROPORTION.  233 

244.    Proposition  I. — In  every  proportion,  the  product 
of  the  means  is  equal  to  the  product  of  the  extremes. 

Let  a  :  b  :  :  c  :  d. 

Then,  since  this  is  a  true  proportion,  we  must  have 

b    d 

Clearing  of  fractions,  we  have  ad=bc. 

Illustration  by  numbers,     3  :  6  :  :  5  :  10,  and  6X5=3x10- 

.  t        <,         ^,    be         ad    ,     ad        , 

From  the  equation  bc=ad,  we  have  a=  — ,  c=-j-,   0= — ,    ana 


be 


a'  b  c 


a— -j,  from  which  we  see,  that  if  any  three  terms  of  a  proportion 
are  given,  the  fourth  may  be  readily  found. 

The  first  three  terms  of  a  proportion,  are  ac,  bd,  and 
acxy;  what  is  the  fourth?  Ans.  bdxy. 


Remark. — This  proposition  furnishes  a  more  convenient  test  of 
the  proportionality  of  four  quantities,  than  the  method  given  in 
Article  240.  Thus,  3  :  8  :  :  2  :  5  is  a  false  proportion,  since  3X^  is 
not  equal  to  8x2. 


245.  Proposition  II. — Conversely,  If  the  product  of 
two  Quantities  is  equal  to  the  product  of  two  others,  two  of 
them  may  be  made  the  means,  and  the  other  two  the  extremes 
of  a  proportion. 

Let  bc=ad. 

Dividing  each  of  these  equals  by  ac,  we  have 

be  __ad  b     d 

ac~  ac'      '  a     c" 

That  is,  a  :  b-  :  C:d. 

Illustration,     5x8=4x10,  and  4  :  5  :  :  8  :  10. 

In  applying  this  Prop.,  take  either  factor  on  either  stde  of  the  equation 
for  the  first  term  of  the  proportion,  pass  to  the  other  side  of  the  equation 
1st  Bk.  20 


234 


RAY'S  ALGEBRA,  FIRST  BOOK. 


for  the  mean  terms,  and  return  for  the  fourth  term.     Eight  proportions 
may  be  written  from  each  of  the  above  equations.     Thus : 

bc=ad  5X8=4X10 


5 

4: 

:10 

5 

10: 

:    4 

8 

4: 

:10: 

8 

10: 

:    4: 

4 

5: 

:    8 

4 

8: 

:     5 

10 

5: 

:     8: 

10 

8- 

:     5: 

246.  Proposition  III. — If  three  quantities  are  in  con- 
tinued proportion,  the  product  of  the  extremes  is  equal  to  the 
square  of  the  mean. 

If 

Then,  by  Art.  244, 


a:  b  :  ■  b  :  C, 
ac  =  bb=b2. 


It  follows,  from  Art.  245,  that  the  converse  of  this  proposition  is 
also  true.     Thus,  if 

ac=b\ 
Then,  a  :  6  :  :  b  :  c ;  that  is, 

If  the  product  of  the  first  and  third  of  two  quantities  is  equal  to  the 
square  of  a  second,  the  first  is  to  the  second  as  the  second  is  to  the  third. 

Illustration:    If  4  :  6  :  :  6  :  9,  then,  4X9=62=36. 

If  2X8=16,  then,  2  :  ^16":  :  /IB":  8. 

Or,  2  :  4  :  :  4  :  8. 


24*7.  Proposition  IV. — If  four  quantities  are  in  pro- 
portion, they  will  be  in  proportion  by  alternation  ;  that 
is,  the  first  will  have  the  same  ratio  to  the  third  that  the  sec- 
ond has  to  the  fourth. 


Let 

This  gives 


a  :  b  •  :  C  :  d. 

bd 

Multiplying  both  sides  by  j.  — v 
That  is,  a  ;  c  :  :  b  :  d. 

Illustration,     2  :  7  :  :  6  :  21,  and  2  :  6  :  :  7  :  21. 


be        a 


RATIO  AND  PROPORTION.  235 

248.  Proposition  V. — If  four  quantities  are  in  propor- 
tion, they  will  be  in  proportion  by  INVERSION  ;  that  is,  the 
second  will  be  to  the  first  as  the  fourth  to  the  third. 


Let                   *                             a  :  b  :  :  G  :  d. 

m                                                                 b      d 

This  gives                                               —  =  — . 

&                                               a      g 

a      g 
Inverting  the  fractions,  we  have        -r  ■=.-=. 

That  is,                                         b  :  a  :  :  d  :  c. 

Illustration,     2  :  5  :  :  6  :  15,  and  5  :  2  :  :  15 

:6. 

249.  Proposition  VI. — If  two  sets  of  proportions  have 
an  antecedent  and  consequent  in  the  one,  equal  to  an  ante- 
cedent and  consequent  in  the  other,  the  remaining  terms  will 
be  proportional. 


a     e 


Let 

a  :  b  : 

:c:d.     (1) 

And 

a  :  b  : 

:e:f.      (2) 

Then  will 

C  :  d  : 

:e:f. 

For,  from  (1), 

b      d 
a~c' 
d_f 

and  from  (] 

Hence, 

c  ~~  e 

This  gives 

c  :d  : 

e:f. 

Illustration, 

3.    5: 

:  6  :  10. 

3:    5: 

:  9  :  15. 

And 

6:10: 

;  9  :  15. 

250.  Proposition  VII. — If  four  quantities  are  in  pro- 
portion, they  will  be  in  proportion  by  composition  ;  that  is, 
the  sum  of  the  first  and  second  will  be  to  the  second,  as  the 
sum  of  the  third  and  fourth  is  to  the  fourth. 

g  :  d. 

b  :  :  c+d :  d. 


Let 

a    b: 

Then  will 

a-\-b  : 

From  the  1st  proportion, 

6  d 
a~G' 

Inverting  the  fractions, 

a  c 
b~ct' 

236  RAY'S  ALGEBRA,  FIRST  BOOK. 


Adding  unity  to  each,        j-  -f-l=-j+l 


Therefore, 


6  '        d 
a-\-b_e-\-d 

This  gives  b  :  a-f-6:  :  d :  e+d; 

Or,  by  inversion,  a-j-6  :  6  :  :  c+d  :  d. 

Illustration,  3  :  4  :  .  6  :  8 

3+4:4:  :  6+8:8; 
Or,  7  ;  4  :  :  14  :  8. 


Remark. — In  a  similar  manner,  it  may  be  proved,  that  the  sum 
of  the  first  and  second  terms  will  be  to  the  first,  as  the  sum  of  the 
third  and  fourth  is  to  the  third. 


251.  Proposition  VIII. — If  four  quantities  are  in  pro- 
portion, they  will  be  in  proportion  by  DIVISION  ;  that  is,  the 
difference  of  the  first  and  second  will  be  to  the  second,  as  the 
difference  of  the  third  and  fourth  is  to  the  fourth. 

Let  a  :  b  :  :  C  :  d. 

Then  will  a—b  :  6  :  :  C— d  :  d. 

From  the  1st  proportion,  —  =s=— . 

01       G 

a       G 

Inverting  the  fractions,  h~7j' 

Ct  G 

Subtracting  unity  from  each,   j- — 1=^ — 1. 


m,       n  a—b      g — d 

Therefore,  — «—  = — 3— 

6  d 

This  gives  6  :  a—b  :  :  d  :  C — d ; 

Or,  by  inversion,  a — 6  :  6  :  :  C — d  :  d. 


Illustration, 

8:5: 

:  16  :  10 

8-5:5: 

:  16—10  :  10 

Or, 

3:5: 

:  6  :  10. 

Remark. — In  a  similar  manner,  it  may  be  proved  that  the  dif- 
ference of  the  first  and  second  will  be  to  the  first,  as  the  difference 
of  the  third  and  fourth  is  to  the  third. 


RATIO  AND  PROPORTION.  237 

252.  Proposition  IX. — If  four  quantities  are  in  pro- 
portion, the  sum  of  the  first  and  second  will  be  to  their 
difference,  as  the  sum  of  the  third  and  fourth  is  to  their 
difference. 

Let  a  :  6  :  :  c  :  d.     (1) 

Then  will  a-\- b  •.  a — b  :  :  c-\-d  :  c— d. 

From  (1),  by  composition,  a-\-b  :  b  :  :  c-\-d  :  d.     (2) 

From  (1),  by  division,  a—b  :  b  :  :  c — d  :  d.     (3) 

By  alternation,  (2)  and  (3)  become  a-\-b  :  c-\-d  :  :  b  :  d. 

a — 6  :  C — d  :  :  b  :  d. 
Therefore,  Art.  249,  a-f-6  :  c+d  :  :  a—b  :  c—d; 

Or,  by  alternation,  a-\-b  :  a — b  :  :  c-{-d  :  c — d. 

Illustration,  5  :  3  :  :  10  :  6. 

5+3  :  5—3  :  :  10+6  :  10     6. 
Or,  8  :  2  :  :  16  :  4. 

253.  Proposition  X. — If  four  quantities  are  in  propor- 
tion, like  powers  or  roots  of  those  quantities  will  also  be  in 
proportion. 


Let 

a  :  b  :  :  C  :  d. 

Then  will 

an  :  bn  :  ■  Gn  :  dn. 

For,  since 

b      d 
a~c 

Raising  to  the  n 

th 

bn      dn 
Power,       -=-. 

That  is,  an:bn::  Cn  :  dn, 

Where  n  may  either  be  a  whole  number  or  a  fraction. 


Illustration, 

2:3: 

:  4  :  6. 

22  :  32  : 

:  42  :  62. 

Or, 

4:9: 

:  16  :  36. 

Also, 

«2  :  62 : 

:  m2a2 :  m2b2. 

And 

:  y/m2a2 :  !/m262. 

Or, 

a.  b\ 

:  ma :  rnb. 

238 


RAY'S  ALGEBRA,  FIRST  BOOK. 


254.  Proposition  XI. — If  two  sets  of  quantities  are  in 
proportion,  the  products  of  the  corresponding  terms  will  also 
be  in  proportion. 


Let 

a  :  6  i 

.C.d.     (1) 

And 

m  :  n  : 

:  r  :  s.      (2) 

Then  will 

am  :  bn  : 

:  cr :  ds. 

For,  from  the  1st, 

b 
~a~ 

d 

And,  from  the  2d, 

n 

771  ~ 

s 
~  r 

Multiplying  these 

,,        bn 

together, = 

1  am 

ds 

~  er 

This  gives 

am  ;  bn  : 

:  cr  :  ds. 

Illustration, 

3:5: 

:  6  :  10. 

4:3: 

:8:6. 

12:15: 

:  48  :  60. 

255.  Proposition  XII. — In  any  continued  proportion, 
that  is,  any  number  of  proportions  having  the  same  ratio,  any 
one  antecedent  is  to  its  consequent,  as  the  sum  of  all  the  ante- 
cedents is  to  the  sum  of  all  the  consequents. 

Let  a  :  b  '.  :  C  :  d  ;  :  m  :  n,  etc. 

Then  will  a  :  b  :  :  a-\-c-\-m  :  b-\-d-\-n. 

a  :  b  :  :  C  :  d, 

a  :  b  :  :  m  :  n, 

bc=ad. 

bm— an. 

ab=ab. 


Since 

And 

We  have 

And 

Add  ab  to  each,  or  put 

The  sums  of  these  equalities  give 

ab  +  bc-\-  b  m—ab-{-  ad-\-  an. 

&(a+c+m)— a(6+d+w). 

a  :b  :  :  a-\-c-\-m  :  6+d+n. 

3  :  4  :  :  6  :  8  :  :  9  :  12. 
3:4::  3+6+9  :  4+8+12. 
Or,  3  :  4  :  :  18  :  24. 


Or, 

Therefore,  Art.  245 

Illustration, 


Remark. — There  are  several  other  propositions  in  Proportion, 
that  may  be  easily  demonstrated,  in  a  manner  similar  to  the  preced- 
ing, but  they  are  of  little  use. 

^8RA^> 

OF  THE 

INIIVETDQITV 


GENERAL  REVIEW.  239 


GENERAL    REVIEW. 

Define  mathematics.  State  points  of  difference  between  arith- 
metic and  algebra.  What  the  unit  of  measure  of  17  bushels? 
Of  5  feet?  Define  a  power.  How  express  a  known  and  unknown 
quantity?  Write  the  principal  signs  used  in  algebra.  Define  a 
residual  quantity.  The  reciprocal  of  a  quantity.  Of  a  fraction. 
What  is  an  algebraic  expression  ? 

Define  addition.  Is  addition  the  same  process  in  algebra  and 
arithmetic?  State  the  general  rule  for  addition.  Define  subtrac- 
tion. In  algebra,  does  the  term  difference  denote  a  number  less  than 
the  minuend?  State  the  rule  for  subtraction.  Define  multiplica- 
tion. State  the  rule  for  the  signs.  For  the  exponents.  Give  gen- 
eral rule  for  multiplication. 

Define  division.  What  is  a  prime  number?  When  are  two  quan- 
tities prime  to  each  other?  Define  the  greatest  common  divisor. 
The  least  common  multiple.  A  fraction.  State  and  illustrate  Prop- 
osition I.;  IV.;  VI.  Rule  for  the  signs  of  fractions  in  multiplica- 
tion. How  resolve  a  fraction  into  an  infinite  series?  Rule  for 
dividing  a  fraction  by  a  fraction. 

What  is  an  equation?  Define  a  quadratic  equation.  A  numeric- 
al equation.  Literal  equation.  How  is  every  equation  to  be  re- 
garded? What  is  solving  an  equation?  State  the  six  axioms. 
Define  an  axiom.  Transposition.  How  clear  an  equation  of  frac- 
tions ?  In  how  many  ways  may  the  unknown  quantity  be  connected 
with  the  known,  and  how  separated  in  each  case?  Rule  for  solution 
of  simple  equations. 

In  the  solution  of  a  problem,  what  are  explicit  conditions?  What 
are  implied  conditions?  Define  elimination.  How  many  and  what 
methods  of  elimination?  Define  and  give  illustration  of  elimination 
by  substitution.  State  the  rule.  By  comparison.  Give  rule.  By 
addition  and  subtraction.  Rule.  How  form  equations  when  the 
problem  contains  three  unknown  quantities? 

What  is  a  negative  solution?  What  does  it  indicate?  State  dif- 
ference between  a  formula  and  a  rule.  What  is  meant  by  generali- 
zation? Define  an  independent  equation.  Illustrate  by  example. 
Define  an  indeterminate  equation.  Illustrate  by  example.  What 
are  redundant  conditions?  Show  that  a  simple  equation  has  but 
one  root. 

Define  power.  Root.  Exponent.  Coefficient.  State  rule  for 
raising  a  monomial  to  any  given  power.  A  polynomial.  A  frac- 
tion. State  the  four  laws  found  by  examining  the  different  powers 
of  a  binomial.     What  the  law  of  the  number  of  terms  in  any  power 


240 

of  a  binomial  ?  Of  the  signs  of  the  terms  ?  Of  the  exponents  of 
the  letters?  Of  the  coefficients  of  the  terms  ?  State  the  uses  of  the 
binomial  theorem. 

Define  evolution.  What  relation  exists  between  the  number  of 
places  of  figures  in  any  number,  and  the  number  of  places  in  its 
square?  Of  what  is  every  number  composed?  State  the  rule  for 
the  extraction  of  the  square  root  of  numbers.  What  is  the  differ- 
ence between  the  squares  of  two  consecutive  numbers  ? 

Define  a  perfect  square.  An  imperfect  square.  A  surd.  How 
extract  the  square  root  of  a  decimal?  Of  a  fraction,  when  both 
terms  are  not  perfect  squares?  Rule  for  extraction  of  the  square 
root  of  a  monomial.  According  to  what  law  is  the  square  of  a  poly- 
nomial formed?  State  rule  for  extracting  the  square  root  of  a 
polynomial.  What  are  radicals  of  the  second  degree  ?  What  are 
similar  radicals? 

Rule  for  the  addition  of  radicals  of  the  second  degree.  For  sub- 
traction. Multiplication.  Division.  Of  a  fraction  whose  denom- 
inator contains  a  radical.  Rule  for  the  solution  of  a  pure  quadratic 
equation.  State  the  difference  between  a  pure  and  an  affected  quad- 
ratic equation.  Rule  for  the  solution  of  an  affected  quadratic  equa- 
tion.    Give  the  Hindoo  rule. 

What  is  a  series?  An  arithmetical  progression?  Give  example 
of  an  increasing  arithmetical  progression.  Of  a  decreasing.  State 
rules  for  both  increasing  and  decreasing  arithmetical  series.  Define 
geometrical  progression.  Illustrate  both  an  increasing  and  decreas- 
ing geometrical  progression.  Rule  for  finding  the  sum  of  a  geomet- 
rical series. 

Define  ratio.  In  how  many  ways  may  quantities  of  the  same 
kind  be  compared?  Illustrate  by  examples.  What  are  the  terms 
of  a  ratio?  By  what  two  methods  is  ratio  expressed?  Define  com- 
pound ratio.  Proportion.  How  is  proportion  indicated  ?  State  the 
difference  between  a  ratio  and  a  proportion.  Give  the  terms  of  a 
proportion.     State  Proposition  I. ;  IV. ;  V. 


RAY'S  HIGHER  ALGEBRA,  SECOND  BOOK. 

RAY'S  ALGEBRA,  SECOND  BOOK,  for  advanced  students,  contains  a  concise 
review  of  the  elementary  principles  presented  in  the  First  Book,  with  more  diffi- 
cult examples  for  practice.  Also,  a  full  discussion  of  the  higher  practical  parts  of 
the  science,  embracing  the  General  Theory  of  Equations,  with  Sturm's  celebrated 
theorem,  illustrated  by  examples  ;  Horner's  method  of  resolving  numerical  equa- 
tions, etc.,  etc.    A  thorough  treatise  for  High  Schools  and  Colleges. 

THE    END. 


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ti   J.    tO\ 

'  * 

AUG  i     1938 

qk/RAIV      0* 

tJJPIS      &&      9  1 

LD  21-95m-7,'37 

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183990 


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